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Full text of "Text-book of theorectical naval architecture"

BY THE SAME AUTHOR 

WAR SHIPS 

A Text-book on the Construction, Protection, 
Stability, Turning, etc., of War Vessels 

With 219 Diagrams and 48 pages of plain paper 
at end for Notes and References 

Medium 8vo, I2S. 6d. net 



A TEXT-BOOK ON LAYING OFF 

or, the Geometry of Shipbuilding 

By EDWARD L. ATTWOOD, M.Inst.N.A., 
R.C.N.C., and I. C. G. COOPER, Senior Lofts- 
man, H.M. Dockyard, Chatham, etc With 
121 Diagrams. 8vo, 6s. net. 

LONGMANS, GREEN AND CO. 

LONDON, NEW YORK, BOMBAY, CALCUTTA AND MADRAS 



TEXT-BOOK 

OF 

THEORETICAL NAVAL 
ARCHITECTURE ,.., 



BY *' '*, :,, ; ...'K:' 
EDWARD L. ATTWOOD, M. INST. N.A. 

MEMBER OF ROYAL CORPS OF NAVAL CONSTRUCTORS 
FORMERLY LECTURER ON NAVAL ARCHITECTURE AT THE ROYAL NAVAL COLLEGI 

GREENWICH 
AUTHOR OF "WAR SHIPS" AND "THE MODERN WAR SHIP" 

JOINT AUTHOR OF " LAYING OFF, OR THE GEOMETRY OF SHIPBUILDING " 




WITH NUMEROUS DIAGRAMS 

NEW IMPRESSION 

LONGMANS, GREEN AND CO, 
39 PATERNOSTER ROW, LONDON 

FOURTH AVENUE & S OTH STREET, NEW YORK 
BOMBAY, CALCUTTA, AND MADRAS 

I9I7 

All rights reserved 




Engineering 
Library 

BIBLIOGRAPHICAL NOTE 

First Printed .... February r , 1899. 

Second Edition .... November ^ 1900. 

Third Edition .... October^ 1902. 

Fourth Edition. . . . September \ 1905. 

Edition .... December -, 1908. 

Edition . . , . February ', 1912. 

Reprinted March> 1915. 

Edition .... y?/-^, 1916. 

Impression . . . fttly, 1917. 



PREFACE 

THIS book has been prepared in order to provide students and 
draughtsmen engaged in Shipbuilders' and Naval Architects 
drawing offices with a text-book which should explain the 
calculations which continually have to be performed. It is 
intended, also, that the work, and more especially its later 
portions, shall serve as a text-book for the theoretical portion 
of the examinations of the Science and Art Department in 
Naval Architecture. It has not been found possible to include 
all the subjects given in the Honours portion of the syllabus, 
such as advanced stability work, the rolling of ships, the vibra- 
tion of ships, etc. These subjects will be found fully treated 
in one or other of the books given in the list on page 488. 

A special feature of the book is the large number of 
examples given in the text and at the ends of the chapters. 
By means of these examples, the student is able to test his 
grasp of the principles and processes given in the text. It is 
hoped that these examples, many of which have been taken 
from actual drawing office calculations, will form a valuable 
feature of the book. 

Particulars are given throughout the work and at the end 
as to the books that should be consulted for fuller treatment of 
the subjects dealt with. 

In the Appendix are given the syllabus and specimen 
questions of the examination in Naval Architecture conducted 

840838 a 3 



vi Preface. 

by the Science and Art Department. These are given by the 
permission of the Controller of His Majesty's Stationery Office. 
I have to thank Mr. A. W. Johns, Instructor in Naval 
Architecture at the Royal Naval College, Greenwich, for 
reading through the proofs and for sundry suggestions. I also 
wish to express my indebtedness to Sir W. H. White, K.C.B., 
F.R.S., Assistant Controller and Director of Naval Construction 
of the Royal Navy, for the interest he has shown and the 
encouragement he has given me during the progress of the 
book. 

E. L. ATTWOOD. 

LONDON, 

February^ 1899. 



PREFACE TO THE NEW 
EDITION. 

IN the present edition the matter has been somewhat re- 
arranged and a number of additions made. Two new chapters 
have been added, one on launching calculations, and one on 
the turning of ships. 

LONDON, 1916. 



REMARKS ON EDUCATION IN 
NAVAL ARCHITECTURE 

FOR the bulk of those who study the subject of Naval Archi- 
tecture, the only instruction possible is obtained in evening 
classes, and this must be supplemented by private study. The 
institutions in which systematic instruction in day courses 
is given are few in number, viz. (i) Armstrong College, 
University of Durham, Newcastle-on-Tyne ; (2) University 
of Glasgow ; (3) University of Liverpool ; (4) Royal Naval 
College, Greenwich ; and students who can obtain the advan- 
tage of this training are comparatively few in number. An 
account of the course at Glasgow is to be found in a paper 
before the I.N.A. in 1889 by the late Prof. Jenkins, and at the 
Royal Naval College, in a paper before the I.N.A. in 1905 
by the writer ; see also a paper by Professor Welch on the 
scientific education of naval architects before N.E. Coast 
Institution, 1909. There are scholarships to be obtained for 
such higher education, particulars of which can be had by 
application to the Glasgow, Liverpool, and Newcastle Colleges, 
to the Secretary of the Admiralty, Whitehall, S.W., and to the 
Secretary of the Institution of Naval Architects, Adelphi 
Terrace, Strand. In these courses it is recognized that the 
study of other subjects must proceed concurrently with that 
of Naval Architecture. 

The Naval Architect has to be responsible for the ship 
as a complete design, and in this capacity should have some 
familiarity with all that pertains to a ship. Thus he should 
know something of Marine Engineering (especially of pro- 
pellers) ; of Electricity and Magnetism ; of armour, guns and 
gun-mountings in warships; of masts, rig, etc., in sailing 



viii Remarks on Education in Naval Architecture. 

vessels ; of the work of the stevedore in cargo vessels ; of 
questions relating to the docking and undocking of ships ; of 
appliances for loading and unloading of ships ; of the regula- 
tions of the Registration Societies and the Board of Trade 
regarding structure, freeboard, and tonnage ; of appliances for 
navigating, as well as having a thorough knowledge of the 
practical work of the shipyard. In the early stages of a design, 
the naval architect frequently has to proceed independently in 
trying alternatives for the desired result, and it is not until the 
design is somewhat matured that he can call in the assistance 
of specialists in other departments. The naval architect 
should, therefore, have an interest in everything connected 
with the type of ship he has to deal with, and he will con- 
tinually be collecting data which may be of use to him in his 
subsequent work. 

For the average student of Naval Architecture, in addition 
to the work he does and observes in the shipyard, mould loft, 
and drawing office, it is necessary to attend evening classes in 
Naval Architecture and other subjects. The apprentice should 
systematically map out his time for this purpose. In the first 
place, a good grounding should be obtained in mechanical 
drawing and in elementary mathematics. Both of these sub- 
jects are now taught by admirable methods. The drawing 
classes are usually primarily intended for Engineering students, 
but this is no drawback, as it will familiarize the student with 
drawings of engineering details which he will find of consider- 
able service to him in his subsequent work. Some institutions 
very wisely do not allow students to take up the study of any 
special subject, as Naval Architecture, until they have proved 
themselves proficient in elementary drawing and mathematics. 
The time thus spent is a most profitable investment. 

The Board of Education now only hold examinations in 
two stages, a " lower " and a "higher," see p. 446, but teachers 
will probably divide the work between these stages, and 
themselves hold examinations. 

We will suppose, then, that a student starts definitely with 
the lowest class in Naval Architecture. With this subject 
he should also take up Elementary Applied Mechanics, and, 



Remarks on Education in Naval Architecture, ix 

if possible, some Mathematics. The next year may be devoted 
to the Board of Education Lower Examination in Naval 
Architecture, with a course in more advanced Applied 
Mechanics, and a course in Magnetism and Electricity or 
Chemistry would form a welcome relief. The next year may 
be devoted to further study in Mathematics, Theoretical and 
Applied Mechanics, Electricity and Magnetism. The next 
year may be devoted to another class in Naval Architecture, 
with more advanced Mathematics, including the Differential 
and Integral Calculus. This latter branch of mathematics is 
essential in order to make any progress in the higher branches 
of any engineering subject. If the student is fortunate enough 
to live in a large shipbuilding district, he will be able to attend 
lectures preparing him for the Board of Education Higher 
Stage Examination in Naval Architecture. A first-class 
certificate in this stage is worth having, and in preparing 
for the examination, the student must to a large extent read 
on his own account, and for that year he will be well advised 
to devote his whole attention to this subject. Much will 
depend on the particular arrangements of teaching adopted 
in a district as to how the work can be best spread over a 
series of years. 

In making the above remarks, the writer wishes to empha- 
size the fact that a student cannot be said to learn Naval 
Architecture by merely attending Naval Architecture classes. 
Teachers in this subject have not the time to teach Geometry, 
Applied Mechanics, or Mathematics, and unless these subjects 
are familiar to the student, his education will be of a very 
superficial nature. Teachers of the subject are always ready 
to advise students as to the course of study likely to be most 
beneficial in any given case. 

Students are strongly advised to make themselves familiar 
with the use of the " slide rule," which enables ship calcula- 
tions to be rapidly performed. 



CONTENTS 

CHAPTER FACE 

I. AREAS, VOLUMES, WEIGHTS, DISPLACEMENT, ETC. . . I 
II. MOMENTS, CENTRE OF GRAVITY, CENTRE OF BUOYANCY, 

DISPLACEMENT TABLE, PLANIMETER, ETC 45 

III. CONDITIONS OF EQUILIBRIUM, TRANSVERSE METACENTRE, 

MOMENT OF INERTIA, TRANSVERSE BM, INCLINING 
EXPERIMENT, METACENTRIC HEIGHT, ETC. ... 90 

IV. LONGITUDINAL METACENTRE, LONGITUDINAL BM, 

CHANGE OF TRIM 144 

V. STATICAL STABILITY, CURVES OF STABILITY, CALCULA- 
TIONS FOR CURVES OF STABILITY, INTEGRATOR, 

DYNAMICAL STABILITY 174 

VI. CALCULATIONS OF WEIGHTS STRENGTH OF BUTT CON- 
NECTIONS, DAVITS, PILLARS, DERRICKS, SHAFT 

BRACKETS 224 

VII. STRAINS EXPERIENCED BY SHIPS CURVES OF LOADS, 
SHEARING FORCE, AND BENDING MOMENT EQUIVA- 
LENT GIRDER, "SMITH" CORRECTION, TROCHOIDAL 

WAVE 258 

VIII. HORSE-POWER, EFFECTIVE AND INDICATED RESISTANCE 
OF SHIPS COEFFICIENTS OF SPEED LAW OF COM- 
PARISON PROPULSION 298 

IX. THE ROLLING OF SHIPS 348 

X. THE TURNING OF SHIPS STRENGTH OF RUDDER HEADS 381 

XI. LAUNCHING CALCULATIONS 400 

APPENDIX A. SUNDRY PROOFS, TCHEBYCHEFF'S AND 
BROWN'S DISPLACEMENT SHEET, ETC., 
AND MISCELLANEOUS EXAMPLES . . 406 

B. TABLES OF LOGARITHMS 431 

SINES, TANGENTS AND COSINES . . . 436 
SQUARES AND CUBES 438 

C. SYLLABUS OF NAVAL ARCHITECTURE 

EXAMINATIONS 446 

D. QUESTIONS AT NAVAL ARCHITECTURE 

EXAMINATIONS 450 

ANSWERS TO QUESTIONS 485 

BIBLIOGRAPHY 488 

INDEX 491 



LIST OF FOLDING TABLES 

At End of Book 

TABLE I. DISPLACEMENT TABLE BY SIMPSON'S RULES. 
PLATE I. SHEER DRAWING OF A TUG. 

TABLE II. DISPLACEMENT TABLE BY JOINT RULES, TCHEBYCHEFF'S 
AND SIMPSON'S. 

TABLES III., IIlA. DISPLACEMENT TABLB, 
TABLE IV. STABILITY TABLR. 



TEXT-BOOK 

OF 

THEORETICAL NAVAL ARCHITECTURE 



CHAPTER I. 



AREAS, VOLUMES, WEIGHTS, DISPLACEMENT, ETC. 

Areas of Plane Figures. 

A Rectangle. This is a four-sided figure having its opposite 
sides parallel to one another and all its angles right angles. 
Such a figure is shown in D. C. 

Fig. i. Its area is the pro- 
duct of the length and the 
breadth, or AB X BC. Thus 
a rectangular plate 6 feet 
long and 3 feet broad will 
contain 



6 x 3 = 1 8 square feet FlG - 

and if of such a thickness as to weigh 1 2\ Ibs. per square foot, 
will weigh 

18 x 12^ = 225 Ibs. 

A Square. This is a particular case of the above, the 
length being equal to the breadth. Thus a square hatch of 
3^ feet side will have an area of 

-,1 * ?1 _ 7. y ?! 49. 

3a * 3a 2*3 4 

= 12^ square feet 



Theoretical Naval Architecture. 



A Triangle. This is a figure contained by three straight 
lines, as ABC in Fig. 2. From the vertex C drop a perpen- 

dicular on to the base AB 
- (or AB produced, if neces- 
sary). Then the area is 
given by half the product 
of the base into the height, 
or 

|(AB x CD) 

If we draw through the 
apex C a line parallel to 
the base AB, any triangle 
having its apex on this line, 
and navmg AB for its base, will be equal in area to the 
tiiar.gle ABC. If more convenient, we can consider either A 
or B as the apex, and BC or AC accordingly as the base. 

Thus a triangle of base 5-5- feet and perpendicular drawn 
from the apex z\ feet, will have for its area 




= 6^ square feet 

If this triangle be the form of a plate weighing 20 Ibs. to 
the square foot, the weight of the plate will be 

ff X20=I2 3 lbs. 

A Trapezoid. This is a figure formed of four straight 

lines, of which two only are 
parallel. Fig. 3 gives such a 
figure, ABCD. 

If the lengths of the parallel 
sides AB and CD are a and b 
respectively, and h is the per- 
pendicular distance between 
them, the area of the trapezoid 
a, B. is g iven b y 

FlG - * \(a + b) X h 

or one-half the sum of the parallel sides multiplied by the 
perpendicular distance between them. 




Areas, Volumes, Weights, Displacement, etc. ' 3 

Example. An armour plate is of the form of a trapeoid with parallel 
sides 8' 3" and 8' 9" long, and their distance apart 12 feet. Find its 
weight if 6 inches thick, the material of the armour plate weighing 490 Ibs. 
per cubic foot. 

First we must find the area, which is given by 

8' 3" + 8' 9" ^ 

L ^ - 1 X 12 square feet = # x 12 

= 102 square feet 

The plate being 6 inches thick = \ foot, the cubical contents of the 
piate will be 

102 x \ - 51 cubic feet 
The weight will therefore be 



= ii'i5 tons 

A Trapezium is a quadrilateral or four-sided figure of 
which no two sides are parallel. 

Such a figure is ABCD (Fig. 4). Its area may be found 
by drawing a diagonal BD 
and adding together the 
areas of the triangles ABD, 
BDC. These both have the 
same base, BD. Therefore 
from A and C drop per- 
pendiculars AE and CF on 
to BD. Then the area of 
the trapezium is given by 

|(AE + CF) X BD 

Example. Draw a trapezium FIG- 4- 

on scale \ inch = I foot, where 

four sides taken in order are 6, 5, 6, and 10 feet respectively, and the 
diagonal from the starting-point 10 feet. Find its area in square feet. 

Ans. 40 sq. feet. 

A Circle. This is a figure all points of whose boundary 
are equally distant from a fixed point within it called the centre. 
The boundary is called its circumference^ and any line from the 
centre to the circumference is called a radius. Any line passing 
through the centre and with its ends on the circumference 
is called a diameter. 




4 Theoretical Naval Architecture. 

The ratio between the circumference of a circle and its 
diameter is called Tr, 1 and TT = 3*1416, or nearly -^ 

Thus the length of a thin wire forming the circumference 
of a circle of diameter 5 feet is given by 

TT x 5 = 5 X 3*1416 feet 

= 15*7080 feet 

or using TT = ^, the circumference = 5 x " 
= i|a = i 5 f feet 

The circumference of a mast 2' 6" in diameter is given by 
2\ X TT feet = f x ^ 



The area of a circle of diameter d is given by 



Thus a solid pillar 4 inches in diameter has a sectional 
area of 



= 1 2| square inches 

A hollow pillar 5 inches external diameter and inch thick 
will have a sectional area obtained by subtracting the area of 
a circle 4^ inches diameter from the area of a circle 5 inches 
diameter 

_ r ( 5 ) 2 N r (^r\ 

"\ 4 ) \ 4 ) 
= 3*73 square inches 

The same result may be obtained by taking a mean 
diameter of the ring, finding its circumference, and multiplying 
by the breadth of the ring. 

Mean diameter = 4f inches 
Circumference = ~ X ^f- inches 

Area = (^ X - 7 -) X square inches 
= 3' 7 3 square inches as before 

1 This is the Greek letter//, and is always used to denote 3*1416, or ? 7 2 
nearly ; that is, the ratio borne by the circumference of a circle to its 
diameter. 



Areas, Volumes, Weights, Displacement, etc. 



5 



Trapezoidal Rule. 1 We have already seen (p. 2) that 
the area of a trapezoid, as ABCD, Fig. 5, is given by 
i(AD + BC)AB, or calling AD, BC, and AB y^ y* and h 
respectively the area is given by 



If, now, we have two trapezoids joined together, as in 



B. 



FIG. 5. 




Fig. 6, having BE = AB, the area of the added part will be 
given by 



The area of the whole figure is given by 

\(y-i +y*)h + ite 4- y*)h = \^(y\ + 

If we took a third trapezoid and joined on in a similar 
manner, the area of the whole figure would be given by 

i*c*- 



Trapezoidal rule for finding the area of a curvilinear figure, 
as ABCD, Fig. 7. 

Divide the base AB into a convenient number of equal 
parts, as AE, EG, etc., each of length equal to h, say. Set up 
perpendiculars to the base, as EF, GH, etc. If we join DF, 
FH, etc., by straight lines, shown dotted, the area required 
will very nearly equal the sum of the areas of the trapezoids 
ADFE, EFHG, etc. Or using the lengths y lt y* etc., as 
indicated in the figure 



Area = h 



+}>* 



1 The Trapezoidal rule is largely used in France and in the United 
States for ship calculations. 



6 Theoretical Naval Architecture. 

In the case of the area shown in Fig. 7, the area will be 
somewhat greater than that given by this rule. If the curve, 
however, bent towards the base line, the actual area would be 
somewhat less than that given by this rule. In any case, the 
closer the perpendiculars are taken together the less will be 
the error involved by using this rule. Putting this rule into 
words, we have 

To find the area of a curvilinear figure, as ABCD, Fig. 7, 
by means of the trapezoidal rule, divide the base into any con- 
venient number of equal parts, and erect perpendiculars to the base 
meeting the curve ; then to the half -sum of the first and last of 
these add the sum of all the intermediate ones ; the result multi- 
plied by the common distance apart will give tJie area required. 




The perpendiculars to the base AB, as AD, EF, are termed 
" ordinates? and any measurement along the base from a given 
starting-point is termed an "abscissa" Thus the point P on 
the curve has an ordinate OP and an abscissa AO when 
referred to the point A as origin. 

Simpson's First Rule. 1 This rule assumes that the 
curved line DC, forming one boundary of the curvilinear area 
ABCD, Fig. 8, is a portion of a curve known as a parabola of 
the second order? In practice it is found that the results given 
by its application to ordinary curves are very accurate, and it is 

1 It is usual to call these rules Simpson's rules, but the first rule 
was given before Simpson's time by James Stirling, in bis " Methodus 
Differentialis," published in 1730. 

2 A "parabola of the second order" is one whose equation referred 
to co-ordinate axes is of the form^ = a, + a^x + a 2 * 2 , where a,, a lt a 2 are 
constants. 



Areas, Volumes, Weights, Displacement, etc. 



this rule that is most extensively used in this country in finding 
the areas of curvilinear figures required in ship calculations. 

Let ABCD, Fig. 8, be a figure bounded on one side by 
the curved line DC, which, as p 

stated above, is assumed to be 
a parabola of the second order. 
AB is the base, and AD and 
BC are end ordinates perpen- 
dicular to the base. 

Bisect AB in E, and draw 
EF perpendicular to AB, meet- 



ing the curve in F. Then the 
area is given by 






FIG. 8. 



|AE(AD + 4EF + BC) 

or using y^y^y^ to represent the ordinates, h the common 
interval between them 



Now, a long curvilinear area * may be divided up into a 
number of portions similar to the above, to each of which the 
above rule will apply. Thus the area of the portion GHNM 
of the area Fig. 7 will be given by 

-te 
and the portion MNCB will have an area given by 

"On 

3 

Therefore the total area will be, supposing all the ordinates 
are a common distance h apart 

~(y\ + 4^2 + 2^3 
o 

Ordinates, as GH, MN, which divide the figure into the 
elementary areas are termed " dividing ordinates" 

Ordinates between these, as EF, KL, OP, are termed 
" intermediate ordinates" 

1 The curvature is supposed continuous. If the curvature changes 
abruptly at any point, this point must be at a dividing ordinate. 



-f 



8 



Theoretical Naval Architecture. 



Notice that the area must have an even number of intervals^ 
or, what is the same thing, an odd number of ordinates^ for 
Simpson's first rule to be applicable. 

Therefore, putting Simpson's first rule into words, we 
have 

Divide the base into a convenient even number of equal parts, 
and erect ordinates meeting the curve. Then to the sum of the end 
ordinates add four times the even ordinates and twice the odd 
ordinates. The sum thus obtained^ multiplied by one-third the 
common distance apart of the ordinates^ will give the area. 

Approximate Proof of Simpson's First Rule. The 
truth of Simpson's first rule may be understood by the following 
approximate proof : 1 

Let DFC, Fig. 9, be a curved line on the base AB, and 
with end ordinates AD, BC perpendicular to AB. Divide AB 
equally in E, and draw the ordinate EF perpendicular to AB. 
Then with the ordinary notation 



Area = - 



by Simpson's first rule. 



Now 



+ 4^ 




divide AB into three equal 
parts by the points G and H. 
Draw perpendiculars GJ and 
HK to the base AB. At F 
draw a tangent to the curve, 
meeting GJ and HK in J and 
K. Join DJ and KC. Now, 
it is evident that the area we 
want is very nearly equal to 
the area ADJKCB. This 
will be found by adding to- 
gether the areas of the trape- 
zoids ADJG, GJKH, HKCB. 

+ GJ)AG 



FIG. 9. 



Area of ADJG = 

GJKH = i(GJ + HK)GH 
HKCB = i(HK + BC)HB 

1 Another proof will be found on p. 77. The mathematical proof will 
be found in Appendix A. 



Areas, Volumes, Weights, Displacement, etc. 



Now, AG = GH = HB = AB = f AE, therefore the total 
area is 



) (AD + 2GJ + 2HK + BC) 



Now, AE = h, and GJ + HK = 2EF (this may be seen at 
once by measuring with a strip of paper), therefore the total 
area is 

^(AD + 4EF + BC) = -(y, + 4^ +.*) 

J O 

which is the same as that given by Simpson's first rule. 

Application of Simpson's First Rule. Example. A curvi- 
linear area has ordinates at a common distance apart of 2 feet, the lengths 
being 1*45, 2*65, 4-35, 6-45, 8*50, 10-40, and 11-85 feet respectively. 
Find the area of the figure in square feet. 

In finding the area of such a curvilinear figure by means of Simpson's 
first rule, the work is arranged as follows : 



Number of 
ordinate. 


Length of 
ordinate. 


Simpson's 
multipliers. l 


Functions of 
ordinates. 


2 


2-65 


4 


io*6o 


3 


4*35 


2 


870 


4 


6-45 


4 


25-80 


1 


8-50 
10-40 


2 

4 


17-00 
41-60 


7 


11-85 


I 


11*85 



117-00 sum of functions 
Common interval = 2 feet 
\ common interval = feet 

area =117x1 = 78 square feet 

The length of this curvilinear figure is 12 feet, and it has 
been divided into an even number of intervals, viz. 6, 2 feet 
apart, giving an odd number of ordinates, viz. 7. We are 
consequently able to apply Simpson's first rule to finding its 
area. Four columns are used. In the first column are placed 
the numbers of the ordinates, starting from one end of the 
figure. In the second column are placed, in the proper order, 
the lengths of the ordinates corresponding to the numbers in 
the first column. These lengths are expressed in feet and 

1 Sometimes the multipliers used are half these, viz. , 2, I, 2, I, 2, $, 
and the result at the end is multiplied by two-thirds the common interval. 



10 



Theoretical Naval Architecture. 



decimals of a foot, and are best measured off with a decimal 
scale. If a scale showing feet and inches is used, then the 
inches should be converted into decimals of a foot ; thus, 
6' g" = 675', and 6' 3^" = 6' 3'. In the next column are placed 
Simpson's multipliers in their proper order and opposite their 
corresponding ordinates. The order may be remembered by 
combining together the multipliers for the elementary area first 
considered 

i 4 i 

i 4 i 

i 4 i 

or 1424241 

The last column contains the product of the length of the 
ordinate and its multiplier given in the third column. These 
are termed the "functions of ordinates" The sum of the 
figures in the last column is termed the " sum of functions of 
ordinates" This has to be multiplied by one-third the common 
interval, or in this case J. The area then is given by 

117 X | = 78 square feet 

Simpson's Second Rule. This rule assumes that the 
curved line DC, forming one boundary of the curvilinear area 

H. 




A E F B 

FIG. 10. 

ABCD, Fig. 10, is a portion of a curve known as "a parabola 
of the third order" 1 

Let ABCD, Fig. 10, be a figure bounded on one side by 
the curved line DC, which, as stated above, is assumed to be 

1 A "parabola of the third order" is one whose equation referred to 
co-ordinate axes is of the form y = a + a^x + a^x* + a,* 8 , where a g , a^ 
a v fl g are constants. 



Areas, Volumes, Weights, Displacement, etc. n 

" a parabola of the third order" AB is the base, and AD and 
BC are end ordinates perpendicular to the base. Divide the 
base AB into three equal parts by points E and F, and draw 
EG, FH perpendicular to AB, meeting the curve in G and H 
respectively. Then the area is given by 

|AE(AD + 3EG + 3 FH + BC) 

or, using y^ y 2 , y 3 , y^ to represent the ordinates, and h the 
common interval between them 

Area = \h(y^ + 372 + 3? + J^) 

Now, a long curvilinear area 1 may be divided into a 
number of portions similar to the above, to each of which the 
above rule will apply. Thus the area of the portion KLCB in 
Fig. 7 will be given by 

1^4 + 3^5 + 3^6+^7) 

Consequently the total area of ABCD, Fig. 7, will be, 
supposing all the ordinates are a common distance h apart 



The ordinate KL is termed a " dividing ordinate" and the 
others, EF, GH, MN, OP, are termed " intermediate ordinates" 
This rule may be approximately proved by a process similar to 
that adopted on p. 8 for the first rule. 2 

Application of Simpson's Second Rule. Example. A cur- 
vilinear area has ordinates at a common distance apart of 2 feet, the 
lengths being 1-45, 2^65, 4*35, 6*45, 8*50, 10-40, and 11-85 feet respectively. 
Find the area of the figure in square feet by the use of Simpson's second rule. 

In finding the area of such a curvilinear figure by means of Simpson's 
second rule, the work is arranged as follows : 



Number of 
ordinate. 


Length of 
ordinate. 


Simpson's 
multipliers. 


Functions of 
ordinates. 


1 
2 


i-45 
2-65 


I 

3 


i*45 

7'95 


3 


4'35 


3 


13-05 


4 


6-45 


2 


1 2 '9O 


5 


8-50 


3 


25-50 


6 


10-40 


3 


31-20 


7 


1 1 '85 


i 


II'85 



103-90 sum of functions 
Common interval = 2 feet 
| common interval = f = f 

103-9 X f = 77-925 square feet 

1 See footnote on p. 7. 

* See Appendix A for the mathematical proof! 



12 Theoretical Naval Architecture. 

This curvilinear area is the same as already taken for an 
example of the application of Simpson's first rule. It will be 
noticed that the number of intervals is 6 or a multiple of 3. 
We are consequently able to apply Simpson's second rule to 
finding the area. The columns are arranged as in the previous 
case, the multipliers used being those for the second rule. 
The order may be remembered by combining together the 
multipliers for the elementary area with three intervals first 
considered 



or i 3 3 2 3 3 i 
For nine intervals the multipliers would be i, 3, 3, 2, 3, 3, 

2, 3, 3, i. 

The sum of the functions of ordinates has in this case to be 
multiplied by f the common interval, or f x 2 = f , and con- 
sequently the area is 

103*9 X f = 77*925 square feet 

It will be noticed how nearly the area as obtained by the 
two rules agree. In practice the first rule is used in nearly all 
cases, because it is much simpler than the second rule and 
quite as accurate. It sometimes happens, however, that we 
only have four ordinates to deal with, and in this case Simp- 
son's second rule must be used. When there are six ordinates, 
neither of the above rules will fit. The following rule gives 
the area : ff . h(\ . y, + 7a + * + y, + y* + I*). This may 
be proved by applying the second rule to the middle four 
ordinates, and the following 5, 8, i rule to the ends. 

To find the Area of a Portion of a Curvilinear Area 
contained between Two Consecutive Ordinates. Such 
a portion is AEFD, Fig. 8. In order to obtain this area, we 
require the three ordinates to the curve y l y?, jv 3 . The curve 
DFC is assumed to be, as in Simpson's first rule, a parabola of 
the second order. Using the ordinary notation, we have 



Area of ADFE = ^(5^1 + 8y 2 - y 3 ) 
Thus, if the ordinates of the curve in Fig. 8 be 8*5, 10-4, 



Areas, Volumes, Weights, Displacement, etc. 13 

11-85 feet, and 2 feet apart, the area of AEFD will be given 

by- 



y^ X 2(5 x 8-5 + 8 x 10*4 11*85) = 18-97 square feet 
Similarly the area of EBCF will be given by 



^ X 2(5 x 11*85 + 8 X 10-4 - 8-5) = 22-32 square feet 

giving a total area of the whole figure as 41*29 square feet. 

Obtaining this area by means of Simpson's first rule, we 
should obtain 41*3 square feet. 1 

This rule is sometimes known as the "five-eight" rule. 

Subdivided Intervals. When the curvature of a line 
forming a boundary of an area, as Fig. n, is very sharp, it is 
found that the distance apart of ordinates, as used for the 
straighter part of the curve, does not give a sufficiently accurate 
result. In such a case, ordinates B 
are drawn at a sub-multiple of 
the ordinary distance apart of 
the main ordinates. 

Take ABC, a quadrant of a 
circle (Fig. n), and draw the 
three ordinates y 2) y^ y^ a dis- 
tance h apart. Then we should 
get the area approximately by 
putting the ordinates through 
Simpson's first rule. Now, the 
curve EFC is very sharp, and 
the result obtained is very far 
from being an accurate one. 
ordinates y\ y". 
given by 




Now put in the intermediate 
Then the area of the portion DEC will be 



or we may write this 



(y 6 = o at end) 



The area of the portion ABED is given by 



1 See Example 25, p. 41. 



14 Theoretical Naval Architecttire. 

or the area of the whole figure 



Thus the multipliers for ordinates one-half the ordinary distance 
apart are ^, 2, ^, and for ordinates one-quarter the ordinary 
distance apart are -J-, i, , i, \. Thus we diminish the 
multiplier of each ordinate of a set of subdivided intervals in 
the same proportion as the intervals are subdivided. Each 
ordinate is then multiplied by its proper multiplier found in 
this way, and the sum of the products multiplied by \ or f the 
whole interval according as the first or second rule is used 
An exercise on the use and necessity for subdivided intervals 
will be found on p. 43. 

Algebraic Expression for the Area of a Figure 
bounded by a Plane Curve. It is often convenient to be 
able to express in a short form the area of a plane curvilinear 
figure. 

In Fig. 12, let ABCD be a strip cut off by the ordinates 

AB, CD, a distance A* apart, 

A# being supposed small. 

Then the area of this strip is 

very nearly 

y X A* 

where y is the length of the 
ordinate AB. If now we 
imagine the strip to become 

indefinitely narrow, the small 

PIG j triangular piece BDE will dis- 

appear, and calling dx the 
breadth of the strip, its area will be 

y X dx 

The area of the whole curvilinear figure would be found if 
we added together the areas of all such strips, and this could 
be written 

fy.J* 

where the symbol / may be regarded as indicating the sum 
of all such strips as y . dx. We have already found that 



Areas, Volumes, Weights, Displacement, etc. 15 



Simpson's rules enable us to find the areas of such figures, 
so we may look upon the expression for the area 

jy.dx 

as meaning that, to find the area of a figure, we take the 
length of the ordinate y at convenient intervals, and put them 
through Simpson's multipliers. The result, multiplied by \ or 
f the common interval, as the case may be, will give the area. 
A familiarity with the above will be found of great service in 
dealing with moments in the next chapter. 

To find the Area of a Figure bounded by a Plane 
Curve and Two Radii. Let OAB, Fig. 13, be such a figure, 
OA, OB being the 
bounding radii. 

Take two points 
very close together on 
the curve PP' ; join OP, 
OP', and let OP = r 
and the small angle 
POF = A0 in circular 
measure. 1 Then OP 
= OP' = r very nearly, 
and the area of the 
elementary portion 



Fio. 13. 




being the length of PP', 

and regarding OPP' as 

a triangle. If now we consider OP, OP' to become in- 

definitely close together, and consequently the angle POP* 

indefinitely small = dO say, any error in regarding PO P' as a 

triangle will disappear, and we shall have 

Area POP' = - . 46 

2 

and the whole area AOB is the sum of all such areas which 
can be drawn between OA and OB, or 



I-.- 



1 See pp. 16 and 90. 



i6 



Theoretical Naval Architecture. 



Now, this exactly corresponds to the algebraic expression 
for the area of an ordinary plane curvilinear figure, viz. 

ly.dx (seep. 15) 
y corresponding to and dx corresponding to dB. Therefore 

divide the angle between the bounding radii into an even 
number of equal angular intervals by means of radii. Measure 
these radii, and treat their half-squares as ordinates of a curve 
by Simpson's first rule, multiplying the addition by \ the 
common angular interval in circular measure. Simpson's second 
rule may be used in a similar manner. 

The circular measure of an angle^- is the number of degrees 

it contains multiplied by -^-, or 0-01745. Thus the circular 
180 

measure of 



9= 2 



3-I4I6 



and the circular measure of 15 is 0-26175. 

Example. To find the area of a figure bounded by a plane curve and 
two radii 90 apart, the lengths of radii 15 apart being o, 2 '6, 5*2, 7*8, 10*5, 
I3'i, 157. 



Angle from 
first radius. 


Length of 
radius. 


Square of 
length. 


Simpson's 
multipliers. 


Functions of 
squares. 





O'O 


O'O 


I 


O'O 


I 5 


2-6 


6-8 


4 


27-2 


30 


5-2 


27-0 


2 


54'0 


45 


7-8 


60-8 


4 


243-2 


60 


10-5 


IIO'2 


2 


220-4 


75 


13-1 


I71'6 


4 


686-4 


900 


157 


246-5 


I 


246-5 



1477-7 sum of 

functions 
Circular measure of 15 = 0-26175 

/. area = 1477-7 X \ X 0-26175 X 

= 64^ square feet nearly 

The process is exactly the same as in Simpson's rule for a plane area 
with equidistant ordinates. To save labour, the squares of the radii are put 
through the proper multipliers, the multiplication by J being performed at 
the end. 



1 See also p. 90. 



Areas, Volumes, Weights, Displacement, etc. 17 

TehebyehefFs Rules. We have discussed above various 
methods that can be employed for determining the area of a 
figure bounded by a curved line. The methods that are most 
largely employed are those known as the " Trapezoidal rule " 
and " Simpson's first rule." The former is used in France and 
America, 1 and the latter is used in Great Britain. The trape- 
zoidal rule has a great advantage in its simplicity, but con- 
siderable judgment is necessary in its use to obtain good 
results. Simpson's first rule is rather more complex, but gives 
exceedingly good results for the areas dealt with in ordinary 
ship calculations. 

In the above rules the spacing of ordinates is constant. A 
rule has been devised for determining the area of a curvilinear 



figure, in which 
the multiplier is 
the same for all 
ordinates when 
these ordinates y 
are suitably / 
placed so that / 
the lengths of / 


f 

s 

10 


^^ 
o 


> 

& 

Ol 


N 


\\ 


only need add- 
ing together 
to obtain the 


8^25 . 
i 

FIG. 


+27*2-* 

ISA. 





area of the figure. This rule is Tchebycheff's rule, 8 and a 
much fewer number of ordinates are required than for 
either the Trapezoidal or Simpson's rules for equally correct 
results. For instance to find the area of a semicircle 10 feet 
radius, using five ordinates. These ordinates are placed in the 
positions shown in Fig. 1 3A. The lengths of these five ordinates 
are found to be 5-54, 9*27, io'o, 9*27, 5-54 feet respectively, and 
all that is needed is to add these lengths together, multiply the 

1 See a paper read before the American Society of Naval Architects in 
1895, by Mr. D. W. Taylor. 

2 See a paper by Mr. C. F. Munday, M.I.N.A., in the Transactions of 
the Institution of Naval Architects for 1899, and a paper by Professor Biles 
in the Transactions of Institute of Engineers and Shipbuilders in Scotland 
for 1899. 



18 



Theoretical Naval Architecture 



result by the length of the figure, and divide by the number of 
ordinates. The area by this rule is therefore 

20 
39-62 X = 158-48 square feet 

The exact area is, of course, 157:08 square feet. Example 
No. 48, Chapter I., gives an illustration of the number of 
ordinates it is necessary to use for such a figure when using 
Simpson's first rule in order to obtain a close approximation 
to the correct area. 

The following table gives the position of ordinates of a 
curve with reference to the middle ordinate for different 
numbers of ordinates : 



No. of 
ordinates 
used. 


Position of ordinates from middle of base in fractions of the half-length of base. 


2 










0-5773 










3 








o 




0-7071 








4 








0-1876 




0-7947 








1 








0-2666 




0-3745 
0-4225 




0-8325 
0-8662 






I 





O'IO26 




0-3239 

0-4062 




0-5297 
0-5938 




0-8839 
0-8974 




9 







0-1679 




0-5288 




0-6010 




0-9116 


10 


0-0838 




0-3127 




0-5000 




0-6873 




0-9162 



The following example will show how the rule is employed 
to find the area of the load water-plane of a ship 600 feet long, 
for which by the ordinary method 21 ordinates would have to 
be used 30 feet apart. By using 10 ordinates, and drawing 
them according to the above table, i.e. the following distance 
from amidships both forward and aft, 25*1, 63-8, 150*0, 206-2, 
274-9 feet respectively, we obtain the following lengths for the 
semi-ordinates, commencing from forward : 3*6, 15*0, 25*1, 
3i'6, 35'i, 35'4> 33'4, 28*8, 22-2, and 11*5 feet respectively. 
These lengths are added together, and the result is multiplied 
by the length of the water-plane and divided by the number of 
ordinates. The result is the half-area of the water-plane, viz. 

600 
2417 X = 14,502 square feet. 



Areas, Volumes ', Weights, Displacement, etc. 19 

The simplification due to this method consists in the fewer 
number of ordinates necessary and the simple process of 
addition that is required when the ordinates are measured off. 

The method of proof of these rules is given in the Appendix. 

Measurement of Volumes. 

The Capacity or Volume of a Rectangular Block 
is the product of the length, breadth, and depth, or, in other 
words, the area of one face multiplied by the thickness. All 
these dimensions must be expressed in the same units. Thus 
the volume of an armour plate 1 2 feet long, 85 feet wide, and 
1 8 inches thick, is given by 

12 x 8J X 4 = 12 x ^ X f = -f 1 =148! cubic feet. 

The Volume of a Solid of Constant Section is the 
area of its section multiplied by its length. Thus a pipe 2 feet 

in diameter and 100 feet long has a section of = " square 



feet, and a volume of ^ X 100 = 2M2 = 3 14^ cubic feet. 

A hollow pillar 7' 6" long, 5 inches external diameter, and 
\ inch thick, has a sectional area of 

3*73 square inches 



or - square feet 
144 

and the volume of material of which it is composed is 



\i44/ * 2 96 

= 0-195 cubic foot 

Volume of a Sphere. This is given by ? . d?, where d 

is the diameter. Thus the volume of a ball 3 inches in dia- 
meter is given by 

TT __ 22 x 27 
6' 27= ~~^~ 

= i4y cubic inches 
Volume of a Pyramid. This is a solid having a base 



20 



Theoretical Naval Architecture. 



in the shape of a polygon, and a point called its vertex not in 
the same plane as the base. The vertex is joined by straight 
lines to all points on the boundary of the base. Its volume is 
given by the product of the area of the base and one-third the 
perpendicular distance of the vertex from the base. A cone is 
a particular case of the pyramid having for its base a figure 
with a continuous curve, and a right circular cone is a cone 
having for its base a circle and its vertex immediately over the 
centre of the base. 

To find the Volume of a Solid bounded by a 
Curved Surface. The volumes of such bodies as this are 
continually required in ship calculation work, the most 
important cases being the volume of the under-water portion 
of a vessel. In this case, the volume is bounded on one side 
by a plane surface, the water-plane of the vessel. Volumes 
of compartments are frequently required, such as those for 
containing fresh water or coal-bunkers. The body is divided 

by a series of planes 
or AO.. spaced equally apart. 

The area of each section 
is obtained by means of 
one of the rules already 
explained. These areas 
are treated as the ordi- 
nates of a new curve, 
which may be run in, 
with ordinates the spac- 
ing of the planes apart. 
FIG. 14. It is often desirable to 

draw this curve with 

areas as ordinates as in Fig. 14, because, if the surface is a fair 
surface, the curve of areas should be a fair curve, and should 
run evenly through all the spots ; any inaccuracy may then be 
detected. The area of the curve of areas is then obtained by 
one of Simpson's rules as convenient, and this area will re- 
present the cubical contents of the body. 

Example. A coal-bunker has sections if 6" apart, and the areas of 
these sections are 98, 123, 137, 135, 122 square feet respectively. Find the 



Areas, Vohimes, Weights, Displacement, etc. 21 

volume of the bunker and the number of tons of coal it will hold, taking 
44 cubic feet of coal to weigh I ton. 



Areas. 


Simpson's 
multipliers. 


Functions of 
areas. 


9 8 


I 


9 8 


I2 3 


4 


492 


137 


2 


274 


135 


4 


540 


122 


I 


122 



1526 sum of functions 

| common interval = g X 175 = ^ 

.*. volume = 1526 x ^ cubic feet 

= 8902 cubic feet, 
and the bunker will hold ^ = 202 tons 

The under-water portion of a ship is symmetrical about the 
fore-and-aft middle-line plane. 

We may divide the volume in two ways 

1. By equidistant planes perpendicular to the middle-line 
plane and to the load water-plane. 

2. By equidistant planes parallel to the load water-plane. 
The volume as obtained by both methods should be the 

same, and they are used to check each other. 

Examples. I. The under-water portion of a vessel is divided by vertical 
sections 10 feet apart of the following areas : 0-3, 227, 48-8, 73-2, 88-4, 
82*8, 587, 26-2, 3*9 square feet. Find the volume in cubic feet. (The 
curve of sectional areas is given in Fig. 15.) 

CURVE OF SECTIONAL AREAS. 




\ 




5, 4: 

FIG 15. 



The number of ordinates being odd, Simpson's first rule can be applied 
as indicated in the following calculation : 



22 



Theoretical Naval Architecture. 



Number of 
section. 


Area of 
section. 


Simpson's 
multipliers. 


Function of 
area. 


I 


0'3 


I 


0'3 


2 


227 


4 


90-8 


3 


48-8 


2 


97-6 


4 


73-2 


4 


292-8 


5 


88-4 


2 


176-8 


6 


82-8 


4 


331*2 


7 


587 


2 


117-4 


8 


26-2 


4 


104-8 


9 


3*9 


1 


3'9 



1 2 1 5 '6 sum of functions 



\ common interval = ^ 

/. volume = 1215-6 X -^ 

= 4052 cubic feet 

2. The under-water portion of the above vessel is divided by planes 
parallel to the load water-plane and I J feet apart of the following areas : 
944, 795, 605, 396, 231, 120, 68, 25, 8 square feet. Find the volume in 
cubic feet. 

The number of areas being odd, Simpson's first rule can be applied, as 
indicated in the following calculation : 



Number of 
water-line. 


Area of 
water-plane. 


Simpson's 
multipliers. 


Function of 
area. 


I 


944 


I 


944 


2 


795 


4 


3180 


3 


605 


2 


1210 


4 


396 


4 


1584 


5 


231 


2 


462 


6 


120 


4 


480 


7 


68 


2 


136 


8 


25 


4 


IOO 


9 


8 


i 


8 



\ common interval = 



8104 sum of functions 



- 

.*. volume = 8104 X 3 

= 4052 cubic feet 

which is the same result as was obtained above by taking the areas of 
vertical sections and putting them through Simpson's rule. 

In practice this volume is found by means of a " displace- 
ment sheet," or by the " planimeter." See Chapter II. and 
Appendix A. 



Areas, Volumes, Weights, Displacement, etc. 23 

Displacement. The amount of water displaced or put 
aside by a vessel afloat is termed her "displacement" This 
may be reckoned as a volume, when it is expressed in cubic 
feet, or as a weight, when it is expressed in tons. It is usual 
to take salt water to weigh 64 Ibs. per cubic foot, and conse- 
quently a jJ4 Q = 35 cubic feet of salt water will weigh one 
ton. Fresh water, on the other hand, is regarded as weighing 
62^ Ibs. per cubic foot, or 36 cubic feet to the ton. 1 The 
volume displacement is therefore 35 or 36 times the weight dis- 
placement, according as we are dealing with salt or fresh water. 

If a vessel is floating in equilibrium in still water ^ the weight 
of water she displaces miist exactly equal tht weight of the vessel 
herself with everything she has on board. 

That this must be true may be understood from the follow- 
ing illustrations : 

1. Take a large basin and stand it in a dish (see Fig. 16). 
Just fill the basin i ' i 

to the brim with I / 

water. Now care- ||||| | | v ' - \ [~~ 

fully place a 
smaller basin into 
the water. It \~" \ x ~~7 

will be found that V ^~- ~-^ S 

some of the water FlG - l6 - 

in the large basin will be displaced, and water will spill over 
the edge of the large basin into the dish below. It is evident 
that the water displaced by the basin is equal in amount to the 
water that has been caught by the dish, and if this water be 
weighed it will be found, if the experiment be conducted ac- 
curately, that the small basin is equal in weight to the water in 
the dish that is, to the water it has displaced. 

2. Consider a vessel floating in equilibrium in still water, and 
imagine, if it were possible, that the water is solidified, main- 
taining the same level, and therefore the same density. If now 
we lift the vessel out, we shall have a cavity left behind which 

1 It is advisable to occasionally test the water at any particular place to 
obtain the density, which may vary at different states of the tide. Thus we 
have at Clydebank the water is 35*87 cubic feet to the ton ; at Dundee the 
water is 1021 ozs. to cubic foot at high water, and 1006 ozs. at low water. 




24 Theoretical Naval Architecture. 

will be exactly of the form of the under-water portion of the 
ship, as Fig. 17. Now let the cavity be filled up with water. 
The amount of water we pour in will evidently be equal to the 
volume of displacement of the vessel. Now suppose that the 
solidified water outside again becomes liquid. The water we 
have poured in will remain where it is, and will be supported 
by the water surrounding it. The support given, first to the 
vessel and now to the water we have poured in, by the sur- 

WATER SURFACE. 




FIG. 17. 

rounding water must be the same, since the condition of the 
outside water is the same. Consequently, it follows that the 
weight of the vessel must equal the weight of water poured 
in to fill the cavity, or, in other words, the weight of the 
vessel is equal to the weight of water displaced. 

If the vessel whose displacement has been calculated on p. 22 
is floating at her L.W.P. in salt water, her total weight will be 
4052 4-35 = 115*8 tons 

If she floated at the same L.W.P. in fresh water, her total weight 
would be 

4052 4-36 = 112^ tons 

It will be at once seen that this property of floating bodies 
is of very great assistance to us in dealing with ships. For, to 
find the weight of a ship floating at a given line, we do not 
need to estimate the weight of the ship, but we calculate out 
from the drawings the displacement in tons up to the given line, 
and this must equal the total weight of the ship. 

Curve of Displacement. The calculation given on p. 22 
gives the displacement of the vessel up to the load-water plane, 
but the draught of a ship continually varies owing to different 
weights of cargo, coal, stores, etc., on board, and it is desirable 



Areas, Volumes, Weights, Displacement, etc. 25 



to have a means of determining quickly the displacement at 
any given draught. From the rules we have already investi- 
gated, the displacement in tons can be calculated up to each 
water-plane in succession. If we set down a scale of mean 
draughts, and set off perpendiculars to this scale at the places 
where each water-plane comes, and on these set off on a con- 
venient scale the displacement we have found up to that water- 
plane, then we should have a number of spots through which we 
shall be able to pass a fair curve if the calculations are correct. 



SCALE FOR DISPLACEMENT. 

|'OCOTONS~ 




FIG. 1 8, 



A curve obtained in this way is termed a " curve of displacement? 
and at any given mean draught we can measure the displace- 
ment of the vessel at that draught, and consequently know at 
once the total weight of the vessel with everything she has on 
board. This will not be quite accurate if the vessel is floating 
at a water-plane not parallel to the designed load water-plane. 
Fig. 1 8 gives a "curve of displacement" for a vessel, and the 
following calculation shows in detail the method of obtaining 
the information necessary to construct it. 



26 



TJieoretical Naval Architecture. 



The areas of a vessel's water-planes, two feet apart, are 
follows : 



L.W.L. 
2 W.L. 
3W.L. 

4 W.L. 

5 W.L. 

6 W.L. 

7 W.L. 



7800 square feet. 

745 
6960 
6290 
5460 
4320 
2610 



The mean draught to the L.W.L. is 14' o",and the displace 
ment below the lowest W.L. is 7 1 tons. 
To find the displacement to the L.W.L. 



Number of 
W.L. 


Area of 
water-plane. 


Simpson's 
multipliers. 


Function of 
area. 


, 


7800 


, 


7,800 


2 


7450 


4 


29,800 


3 


6960 


2 


13,920 


4 


6290 


4 


25,160 


I 


5460 
4320 


2 

4 


10,920 
I 7 ,280 


7 


26lO 


I 


2,6lQ 



107,490 



\ common interval = \ X 2 
/. displacement in cubic feet = 107,490 X 
and displacement in tons, salt water = 107,490 X 

= 2047 tons without the 
appendage 

Next we require the displacement up to No. 2 W.L., and 
we subtract from the total the displacement of the layer between 
i and 2, which is found by using the five-eight rule as follows : 



Number of 
W.L. 


Area of 
water-plane. 


Simpson's 
multipliers. 


Function of 
area. . 


I 


7800 


5 


39,000 


2 


7450 


8 


59,600 


3 


6960 


i 


-6,960 



91,640 



Areas, Volumes, Weights, Displacement, etc. 27 

Displacement in tons between I A 2 x 

No. i and No. 2 W.L.'s f = 9'> 6 4 * TW.X A 

= 436 tons nearly 

/. the displacement up to No. 2 ) 

,,, T . , > = 1611 tons without the 

W.L. is 2047 - 436 I 

appendage 

The displacement between i and 3 W.L.'s can be found by 
putting the areas of i, 2 and 3 W.L.'s through Simpson's first 
rule, the result being 848 tons nearly. 

.*. the displacement up to No. 
W.L. is 2047 - 848 



"" l S Wlth Ut 

appendage 



The displacement up to No. 4 W.L. can be obtained by 
putting the areas of 4, 5, 6, and 7 W.L.'s through Simpson's 
second rule, the result being 

819 tons without the appendage 

The displacement up to No. 5 W.L. can be obtained by 
putting the areas of 5, 6, and 7 W.L.'s through Simpson's first 
rule, the result being 

482 tons without the appendage 

The displacement up to No. 6 W.L. can be obtained by 
means of the five-eight rule, the result being 

201 tons without the appendage 

Collecting the above results together, and adding in the 
appendage below No. 7 W.L., we have 



Disp] 


acement 


up to L.W.L. 
2 W.L. 

4 W!L! 

5 W.L. 
6 W.L. 
7 W.L. 




2118 to 
1682 
1270 
890 

553 
272 


ns. 



These displacements, set out at the corresponding draughts, 
are shown in Fig. 18, and the fair curve drawn through forms 
the "curve of displacement" of the vessel. It is usual to com- 
plete the curve as indicated right down to the keel, although 



28 



Theoretical Naval Architecture. 



che ship could never float at a less draught than that given by 
the weight of her structure alone, or when she was launched. 

Tons per Inch Immersion. It is frequently necessary 
to know how much a vessel will sink, when floating at a given 
water-line, if certain known weights are placed on board, or 
how much she will rise if certain known weights are removed. 
Since the total displacement of the vessel must equal the weight 
of the vessel herself, the extra displacement caused by putting 
a weight on board must equal this weight. If A is the area 



TONS PER INCH IMMERSION. 

I 10 -, 




FIG. 19. 



of a given water-plane in square feet, then the displacement 
of a layer i foot thick at this water-plane, supposing the vessel 
parallel-sided in its neighbourhood, is 
A cubic feet . 

or tons in salt water 
35 

For a layer i inch thick only, the displacement is 

A 



35 X 12 



tons 






Areas, Volumes, Weights, Displacement, etc. 29 

and this must be the number of tons we must place on board 
in order to sink the vessel i inch, or the number of tons we 
must take out in order to lighten the vessel i inch. This is 
termed the " tons per inch immersion " at the given water-line. 1 
This assumes that the vessel is parallel-sided at the water-line 
for the depth of i inch up and i inch down, which may, for all 
practical purposes, be taken as the case. If, then, we obtain 
the tons per inch immersion at successive water-planes parallel 
to the load water-plane, we shall be able to construct a " curve 
of tons per inch immersion" in the same way in which the curve 
of displacement was constructed. Such a curve is shown in 
Fig. 19, constructed for the same vessel for which the displace- 
ment curve was calculated. By setting up any mean draught, 
say 1 1 feet, we can measure off the " tons per inch immersion," 
supposing the vessel is floating parallel to the load water-plane ; 
in this case it is 17-5- tons. Suppose this ship is floating at a 
mean draught of n feet, and we wish to know how much she 
will lighten by burning 100 tons of coal. We find, as above, 
the tons per inch to be 17^, and the decrease in draught is 

therefore 

ioo-^i7i= 5| inches nearly 

Curve of Areas of Midship Section. This curve is 
sometimes plotted off on the same drawing as the displacement 
curve and the curve of tons per inch immersion. The ordi- 
nates of the immersed part of the midship section being known, 
we can calculate its area up to each of the water-planes in 
exactly the same way as the displacement has been calculated. 
These areas are set out on a convenient scale at the respective 
mean draughts, and a line drawn through the points thus 
obtained. If the calculations are correct, this should be a fair 
curve, and is known as " the curve of areas of midship section" 
By means of this curve we are able to determine the area of 
the midship section up to any given mean draught. 

Fig. 20 gives the curve of areas of midship section for the 
vessel for which we have already determined the displacement 
curve and the curve of tons per inch immersion. 

Coefficient of Fineness of Midship Section. If we 

1 For approximate values of the "tons per inch immersion" in various 
types of ships, see Example 55, p. 44. 



30 Theoretical Naval Architecture. 

draw a rectangle with depth equal to the draught of water at 
the midship section to top of keel, and breadth equal to the 



AREAS OF MID; SEC: SQI FT: 

1*00. .300. ,200. 100. 




FIG 20 



extreme breadth at the midship section, we shall obtain what 
may be termed the circumscribing rectangle of the immersed 
midship section. The area of the immersed midship section 
will be less than the area of this rectangle, and the ratio 

area of immersed midship section 
area of its circumscribing rectangle 

is termed the coefficient of fineness of midship section. 

Example. The midship section of a vessel is 68 feet broad at its 
broadest part, and the draught of water is 26 feet. The area of the immersed 
midship section is 1584 square feet. Find the coefficient of fineness of the 
midship section. 

Area of circumscribing rectangle = 68 X 26 

= 1768 square feet 
.'. coefficient = |^|| = 0*895 

If a vessel of similar form to the above has a breadth at 



Areas t Volumes, Weights, Displacement, etc. 3 1 

the midship section of 59' 6" and a draught of 22' 9", the area 
of its immersed midship section will be 

59^ x 22f x 0-895 = I2I 3 square feet 

The value of the midship section coefficient varies in 
ordinary ships from about 0-85 to 0-95, the latter value being 
for a section with very full section. 

Coefficient of Fineness of Water-plane. This is 
the ratio between the area of the water-plane and its circum- 
scribing rectangle. 

The value of this coefficient for the load water-plane may 
be taken as follows : 

For ships with fine ends O'7 

For ships of ordinary form 0*75 

For ships with bluff ends 0*85 

Block Coefficient of Fineness of Displacement. 

This is the ratio of the volume of displacement to the volume 
of a block having the same length between perpendiculars, 
extreme breadth, and mean draught as the vessel. The 
draught should be taken from the top of keel. 

Thus a vessel is 380 feet long, 75 feet broad, with 27' 6" 
mean draught, and 14,150 tons displacement. What is its 
block coefficient of fineness or displacement ? 

Volume of displacement = 14,150 X 35 cubic feet 
Volume of circumscribing solid = 380 X 75 X 27-^ cubic feet 
.'. coefficient of fineness of 1 14150 X 35 

displacement I ~~ 380 x 75 X 27^ 

= 0*63 

This coefficient gives a very good indication of the fineness 
of the underwater portion of a vessel, and can be calculated 
and tabulated for vessels with known speeds. Then, if in the 
early stages of a design we have the desired dimensions given, 
with the speed required, we can select the coefficient of fineness 
which appears most suitable for the vessel, and so determine 
very quickly the displacement that can be obtained under the 
conditions given. 



3 2 Theoretical Naval Architecture. 

Example. A vessel has to be 400 feet long, 42 feet beam, 1 7 feet draught, 
and 13 \ knots speed. What would be the probable displacement ? 

From available data, it would appear that a block coefficient of fineness 
of 0^625 would be desirable. Consequently the displacement would be 

(400 X 42 X 17 X 0*625) * 35 tons =5ioo tons about 

The following may be taken as average values of the block 
coefficient of fineness of displacement in various types of 
ships : 



Recent battleships 
Recent fast cruisers 
Fast mail steamers 
Ordinary steamships 
Cargo steamers 
Sailing vessels 
Steam-yachts 



'6O--65 
*5o-'55 
'5o-*6o 

'5S~'^5 
'65-' 80 

'^5~'7S 
*35-*45 



Prismatic Coefficient of Fineness of Displace- 
ment. This coefficient is often used as a criterion of the 
fineness of the underwater portion of a vessel. It is the ratio 
between the volume of displacement and the volume of a 
prismatic solid the same length between perpendiculars as the 
vessel, and having a constant cross-section equal in area to the 
immersed midship section. 

Example. A vessel is 300 feet long, 2100 tons displacement, and has 
the area of her immersed midship section 425 square feet. What is her 
prismatic coefficient of fineness ? 

Volume of displacement = 2100 X 35 cubic feet 
Volume of prismatic solid = 300 x 425 

x 

0-577 

Difference in Draught of Water when floating 
in Sea Water and when floating in River Water. 
Sea water is denser than river water ; that is to say, a given 
volume of sea water say a cubic foot weighs more than the 
same volume of river water. In consequence of this, a vessel, 
on passing from the river to the sea, if she maintains the same 
weight, will rise in the water, and have a greater freeboard 
than when she started. Sea water weighs 64 Ibs. to the cubic 
foot, and the water in a river such as the Thames may be 



Areas, Volumes ; Weights, Displacement, etc. 33 

taken, as weighing 63 Ibs. to the cubic foot. 1 In Fi^. ,3,1, let 
the right-hand portion represent the ship floating in river water, 
and the left-hand portion represent the ship floating in -salt //>#& 
water. The distance between the two water-planes will be the 
amount the ship will rise on passing into sea water. 



w 


L 




W 


1 






' ' 


W 


L' 


V 


A. J 




^ 1 


B J 



SEA WATER. 



FIG. 21. 

Let W = the weight of the ship in tons ; 

T = the tons per inch immersion at the water-line 

W'L' in salt water ; 
/ = the difference in draught between the water-lines 

WL, W'L' in inches. 
Then the volume of displacement 

W x 2240 
in river water = -- - 

63 



in sea water 



, 

/. the volume of the layer = 



W X 2240 
~~6^~ 

w x 2240 w x 2*40 



63 

W X 2240 






63 X 64 
Now, the volume of the layer also = / x T x ^f^. ; there- 



fore we hav 



22 4 



W 



2240 _ 

-TT- 



1 See also Example 56, p. 44. 



34 



Theoretical Naval Architecture. 



This may be put in another way. A ship, if floating in river 
water, will weigh -^ less than if floating to the same water-line 
in salt water. Thus, if W is the weight of the ship floating at a 
given line in salt water, her weight if floating at the same line 
in river water is 

e^W less 

and this must be the weight of the layer of displacement 
between the salt-water line and the river-water line for a given 
weight W of the ship. If T be the tons per inch for salt water, 
the tons per inch for river water will be ff T. Therefore the. 
difference in draught will be 



-4- |f T 



W 

inches, as above 



Sinkage caused by a Central Compartment of a 
Vessel being open to the Sea. Take the simple case of a 
box-shaped vessel, ABCD, Fig. 22, floating at the water-line WL. 



A. E. G. 


B: 


W 


M 




N. 


L' 




W. K 




R 


L 


" 


3. F 1 
FIG. aa. 


H. 


C 



This vessel has two water-tight athwartship bulkheads in the 
middle portion, EF and GH. A hole is made in the bottom or 
side below water somewhere between these bulkheads. We 
will take a definite case, and work it out in detail to illustrate 
the principles involved in such a problem. 



Length of box-shaped vessel 

Breadth 

Depth 

Draught 

Distance of bulkheads apart 



100 feet 
20 
20 
10 
20 , 



If the vessel is assumed to be floating in salt water, its 
weight must be 

100 x 20 x 10 

tons 



Areas, Volumes, Weights, Displacement, etc. 35 

Now, this weight remains the same after the bilging as 
before, but the buoyancy has been diminished by the opening 
of the compartment KPHF to the sea. This lost buoyancy 
must be made up by the vessel sinking in the water until the 
volume of displacement is the same as it originally was. 
Suppose W'L' to be the new water-line, then the new volume of 
displacement is given by the addition of the volumes of W'MFD 
and NL'CH, or, calling d the new draught of water in feet 

(40 x 20 x d)+ (40 X 20 xd) = 1600^ cubic feet 
The original volume of displacement was 

100 x 20 X 10 = 20,000 cubic feet 
.*. 1600 d = 20,000 

= 12' 6" 



that is, the new draught of water is 12' 6", or the vessel will 
sink a distance of 2' 6". 

The problem may be looked at from another point of view. 
The lost buoyancy 1320X20X10 cubic feet = 4000 cubic 
feet; this has to be made up by the volumes W'MKW and 
NL'LP, or the area of the intact water-plane multiplied by 
the increase in draught. Calling x the increase in draught, we 
shall have 

80 X 20 X x = 4000 



= 2' 6" 

which is the same result as was obtained above. 

If the bilged compartment contains stores, etc., the amount 
of water which enters from the sea will be less than if the com- 
partment were quite empty. The volume of the lost displace- 
mt will then be given by the volume of the compartment up 
the original water-line less the volume occupied by the 
tores. 

Thus, suppose the compartment bilged in the above 
imple to contain coal, stowed so that 44 cubic feet of it will 
reigh one ton, the weight of the solid coal being taken at 
Ibs. to the cubic foot. 



^ Theoretical Naval Architecture. 

i cubic foot of coal, if solid, weighs 80 Ibs. 

i as stowed ^-f-f- = 51 Ibs. 

Therefore in every cubic foot of the compartment there is 

f cubic feet solid coal 

ff- space into which water will find its way 

The lost buoyancy is therefore 

-|f X 4000 =1450 cubic feet 

The area of the intact water-plane will also be affected in 
the same way ; the portion of the water-plane between the bulk- 
heads will contribute 

f X 20 x 20 = 255 square feet to the area 

The area of the intact waterplane is therefore 
1600 -1-255 =185 5 square feet 
The sinkage in feet is therefore 

fHJ=o-78, or 9-36 inches 

In the case of a ship the same principles apply, supposing 
the compartment to be a central one, and we have 

Sinkage of vessel ) _ volume of lost buoyancy in cubic feet 
in feet 1 ~~ area of intact water-plane in square feet 

In the case of a compartment bilged which is not in the 
middle of the length, change of the trim occurs. The method 
of calculating this for any given case will be dealt with in 
Chapter IV. 

In the above example, if the transverse bulkheads EF and 
GH had stopped just below the new water-line W'L', it is 
evident that the water would flow over their tops, and the 
vessel would sink. But if the tops were connected by a water- 
tight flat, the water would then be confined to the space, and 
the vessel would remain afloat. 



Areas, Volumes, Weights, Displacement, etc. 37 

Velocity of Inflow of Water into a Vessel on 
Bilging. 

Let A. = area of the hole in square feet ; 

d = the distance' the centre of the hole below the 

surface in feet ; 

v = initial rate of inflow of the water in feet per 
second. 

Then v %J d nearly 
and consequently the volume of water 
passing through the hole per second 

Thus, if a hole 2 square feet in area, 4 feet below the water- 
line, were made in the side of a vessel, the amount of water, 
approximately, that would flow into the vessel would be as 
follows : 

Cubic feet per second = 8 X /v/4 X 2 

= 3 2 
Cubic feet per minute = 32 x 60 

Tons of water per minute = 

35 



A cub. ft. 



Weights of Materials. The following table gives 
average weights which may be used in calculating the weights 
of materials employed in shipbuilding : 

Steel 490 Ibs. per cubic foot. 



Wrought iron 
Cast iron ... 
Copper 
Brass 


480 
445 
550 


Zinc 




Gunmetal 
Lead 


... 528 
712 


Elm (English) 
(Canadian) 
Fir (Dantzic) 
Greenheart 
Mahogany 
(for boats) 
Oak (English) 
,, (Dantzic) 
(African) 
Pine (Pitch) 
(red) ... 
(yellow) 
Teak ... 


35 
45 
36 
72 
... 40-48 
35 
52 
47 
62 

40 
36 
30 



Theoretical Naval Architecture. 



It follows, from the weights per cubic foot of iron and 
steel given above, that an iron plate i inch thick weighs 40 Ibs. 
per square foot, and a steel plate i inch thick weighs 40*8 Ibs. 
per square foot. 

The weight per square foot may be obtained for other 
thicknesses from these values, and we have the following : 



Thickness in 


Weight per square foot in 
pounds. 




Iron. 


Steel. 


\ 


10 


10-2 


\ 


15 


I5'3 


4 


20 


20- 4 


8 


25 


25'5 




30 


30-6 


i 
i 


35 
40 


357 
40-8 



It is convenient to have the weight of steel per square foot 
when specified in one-twentieths of an inch, as is the case in 
Lloyd's rules 



Thickness in 
inches. 


Weight per 
square foot 
in pounds. 


Thickness in 
inches. 


Weight per 
square foot 
in pounds. 


A 


2-04 


U 


22'44 


A 


4'08 





24-48 


i 


6-12 


1 


26*52 


4 


8-16 




28-56 


& \ 


10-20 


i$ = \ 


30-60 





I2'24 


if 


32-64 


A 


14-28 


kl 


34-68 


A 


16-32 


8 


3672 


1=1 


18-36 

20*40 


1! = i 


3876 
40-80 



Areas, Volumes, Weights, Displacement, etc. 39 



EXAMPLES TO CHAPTER I. 



What is its weight if its 
Ans. 95 Ibs. 




*" IG - 2 3- 



1. A plate has the form shown in Fig. 23. 
weight per square foot is 10 Ibs. ? 

2. The material of an 
armour plate weighs 490 Ibs. 
a cubic foot. A certain 
plate is ordered 400 Ibs. per 
square foot : what is its 
thickness ? 

Ans. 9 '8 inches. 

3. Steel armour plates, 
as in the previous question, 
are ordered 400 Ibs. per 
square foot instead of 10 
inches thick. What is the 
saving of weight per 100 
square feet of surface of this 
armour ? 

Ans. 833 Ibs., or 0-37 ton. 

4. An iron plate is of the dimensions shown in Fig. 24. What is its area ? 
If two lightening holes 2' 3" in diameter are cut in it, what will its 

area then be ? . 

Ans. 33! square feet ; i 

25-8 square feet. . , j 

5. A hollow pillar is 4 inches L _____ ^ 
external diameter and jj inch 

thick. What is its sectional 
area, and what would be the 
weight in pounds of 10 feet of 
this pillar if made of wrought 
iron ? 

Ans. 4-27 square inches ; 
142 Ibs. 

6. A steel plate is of the 

form and dimensions shown in Fig. 25. What is its weight ? (A steel plate 
\ inch thick weighs 25*5 Ibs. per square foot.) 

Ans. 1267 Ibs. 



\ 


A 4 


\ 


M 


rd 



FlG . 




FIG. 25. 

7. A wrought-iron armour plate is 15' 3' 
thick. Calculate its weight in tons. 



long, 3' X 6" wide, and4i inches 
Ans. 4-29 tons. 



4 Theoretical Naval Architecture. 

8. A solid pillar of iron of circular section is 6' 10" long and 2\ inches 
in diameter. What is its weight ? 

Ans. 90^ Ibs. 

9. A Dantzic fir deck plank is 22 feet long and 4 inches thick, and 
tapers in width from 9 inches at one end to 6 inches at the other. What is 
its weight ? 

Ans. 165 Ibs. 

10. A solid pillar of iron is 7' 3" long and 2f inches diameter. What 
is its weight ? 

Ans. 143 Ibs. 

11. The total area of the deck plan of a vessel is 4500 square feet. 
What would be the surface of deck plank to be worked, if there are 

4 hatchways, each 4' X 2$' 
2 ,, ,, 10' X 6' 

and two circular skylights, each 4 feet in diameter, over which no plank is 
to be laid ? 

Ans. 4314-86 square feet. 

12. A pipe is 6 inches diameter inside. How many cubic feet of water 
will a length of 100 feet of this pipe contain? 

Ans. 1 9 '6 cubic feet. 

13. A mast 90 feet in length and 3 feet external diameter, is composed of 
20 Ib. plating worked flush-jointed on three T-bars, each 5" x 3" X iS^lbs. 
per foot. Estimate the weight, omitting straps, and rivet heads. 

Ans. 9^ tons nearly. 

14. A curve has the following ordinates, 1*4" apart: IO'86, 13*53, I4'58, 
15*05, 15*24, 15*28, 15*22 feet respectively. Draw this curve, and find 
its area 

(1) By Simpson's first rule ; 

(2) By Simpson's second rule. 

Ans. (i) 116-07 square feet; (2) 116*03 square feet. 

15. The semi-ordinates in feet of a vessel's midship section, starting 
from the load water-line, are 26*6, 26*8, 26*8, 26*4, 25-4, 23-4, and 18*5 feet 
respectively, the ordinates being 3 feet apart. Below the lowest ordinate 
there is an area for one side of the section of 24*6 square feet. Find the 
area of the midship section, using 

(1) Simpson's first rule ; 

(2) Simpson's second rule. 

Ans. (i) 961 square feet ; (2) 9607 square feet. 

16. The internal dimensions of a tank for holding fresh water are 
8' o" X 3' 6" X 2' 6". How many tons of water will it contain ? 

Ans. I '94. 

17. The yfo^-ordinates of a deck plan in feet are respectively ij, 5^, 
10}, 13^, 14!, 14!, I2j, 9, and 3^, and the length of the plan is 128 feet. 

Find the area of the deck plan in square yards. 

Ans. 296. 

1 8. Referring to the previous question, find the area in square feet of the 
portion of the plan between the ordinates ij and 5^. 

Ans. 106*7. 

19. The half-ordinates of the midship section of a vessel are 22*3, 22*2, 
217, 20'6, 17*2, 13*2, and 8 feet in length respectively. The common 
interval between consecutive ordinates is 3 feet between the first and fifth 
ordinates, and i' 6" between the fifth and seventh. Calculate the total area 
of the section in square feet. 

Ans. 586'2 square feet. 



Areas, Volumes, Weights, Displacement, etc. 4 1 



20. Obtain the total area included between the first and fourth ordinates 
of the section given in the preceding question. 

Ans. 392 '8 square feet. 

21. The semi-ordinates of the load water-plane of a vessel are 0*2, 3-6, 
7-4, 10, n, 10-7, 9-3, 6'5, and 2 feet respectively, and they are 15 feet 
apart. What is the area of the load water -plane ? 

Ans. 1808 square feet. 

22. Referring to the previous question, what weight must be taken out 
of the vessel to lighten her 3 inches ? 

What additional immersion would result by placing 5 tons on board ? 

Ans. 15 tons ; I'i6 inch. 

23. The " tons per inch immersion " of a vessel when floating in salt 
water at a certain water-plane is 44' 5. What is the area of this plane ? 

Ans. 18,690 square feet. 

24. A curvilinear area has ordinates 3 feet apart of length 97, 10*0, and 
13*3 feet respectively. Find 

(1) The area between the first and second ordinates. 

(2) The area between the second and third ordinates. 

(3) Check the addition of these results by finding the area of the whole 

figure by Simpson's first rule. 

25. Assuming the truth of the five-eight rule for finding the area between 
two consecutive ordinates of a curve, prove the truth of the rule known as 
Simpson's first rule. 

26. A curvilinear area has the following ordinates at equidistant intervals 
of 18 feet : 6'2O, 13*80, 21-90, 26*40, 22*35, I 4'7 O > and 7-35 feet. 
Assuming that Simpson's first rule is correct, find the percentage of error 
that would be involved by using 

(1) The trapezoidal rule ; 

(2) Simpson's second rule. 

Ans. (i) i'2 per cent. ; (2) 0*4 per cent. 

27. A compartment for containing fresh water has a mean section of 
the form shown in Fig. 26. The length 9 

of the compartment is 12 feet. How many 8-8. 

tons of water will it contain ? 

Ans. 17 tons. 

28. A compartment 20 feet long, 20 
feet broad, and 8J feet deep, has to be 
lined with teak 3 inches in thickness. 
Estimate the amount of teak required in 
cubic feet, and in tons. 

Ans. 365 cubic feet; 8"6 tons. 

29. The areas of the water-line sec- 
tions of a vessel in square feet are re- 
spectively 2000, 2000, 1600, 1250, and 
300. The common interval between them 
is ij foot. Find the displacement of the 
vessel in tons in salt water, neglecting the 

small portion below the lowest water-line FIG. a6. 

section. 

Ans. 264! tons. 

30. A series of areas, 17' 6" apart, contain 0^94, 2*08, 3*74, 5'33, 8-27, 
12-14, 16-96, 21-82, 24-68, 24-66, 22-56, 17-90, 12-66, 8-40, 5-69, 3-73, 
2 "6 1, 2 "06, o square feet respectively. Find the volume of which the above 
are the sectional areas. 

Ans. 3429 cubic feet. 




4 2 Theoretical Naval Architecture. 

31. Show how to estimate the change in the mean draught of a vessel in 
going from salt to river water, and vice versd. 

A vessel floats at a certain draught in river water, and when floating in 
sea water without any change in lading, it is found that an addition of 175 
tons is required to bring the vessel to the same draught as in river water 
What is the displacement after the addition of the weight named ? 

Ans. 11,200 tons. 

32. The vertical sections of a vessel 10 feet apart have the following 
areas : 10, 50, 60, 70, 50, 40, 20 square feet. Find the volume of displace- 
ment, and the displacement in tons in salt and fresh water. 

Ans. 2966 cubic feet ; 84*7 tons, 82*4 tons. 

33. A cylinder is 500 feet long, 20 feet diameter, and floats with, the 
axis in the water-line. Find its weight when floating thus in salt water. 
What weight should be taken out in order that the cylinder should float 
with its axis in the surface if placed into fresh water ? 

Ans. 2244 tons ; 62 tons. 

34. A vessel is 500 feet long, 60 feet broad, and floats at a mean draught 
of 25 feet when in salt water. Make an approximation to her draught 
when she passes into river water. (Coefficient of displacement. O'5 : coefficient 
ofL.W.P., 0'6.) 

Ans. 25' 4". 

35. A piece of teak is 20 feet long, 4$ inches thick, and its breadth 
tapers from 12 inches at one end to 9 inches at the other end. What is its 
weight, and how many cubic feet of water would it displace if placed into 
fresh water (36 cubic feet to the ton) ? 

Ans. 348 Ibs. ; 5$ cubic feet about. 

36. The area of a water-plane is 5443 square feet. Find the tons per 
inch immersion. Supposing 40 tons placed on board, how much would the 
vessel sink ? 

State any slight error that may be involved in any assumption made. If 
40 tons were taken out, would the vessel rise the same amount ? What 
further information would you require to give a more accurate answer ? 

Ans. 12*96 tons; 3*1 inches nearly. 

37. Bilge keels are to be fitted to a ship whose tons per inch is 48. 
The estimated weight of the bilge keels is 36 tons, and the volume they 
occupy is 840 cubic feet. What will be the increase of draught due to 
fitting these bilge keels? 

Ans. \ inch. 

38. The tons per inch of a vessel at water-lines 2 feet apart are 19-45, 
18-51, 17-25, 15-6, 13-55, 10-87, and 6-52, the lowest water-line being 18 
inches above the underside of flat keel. Draw the curve of tons per inch 
immersion to scale, and estimate the number of tons necessary to sink the 
vessel from a draught of 12 feet to a draught of 13' 6". 

Ans. 344 tons. 

39. The steamship Umbria is 500 feet long, 57 feet broad, 22' 6" 
draught, 9860 tons displacement, 1 150 square feet area of immersed midship 
section. Find 

(1) Block coefficient of displacement. 

(2) Prismatic ,, ,, 

(3) Midship-section coefficient. 

Ans. (i) 0-538; (2) 0-6 ; (3) 0-896. 

40. The steamship Orient is 445 feet long, 46 feet broad, 21' 4^" draught 
mean ; the midship section coefficient is 0-919, the block coefficient of dis- 
placement is 0-621. Find 



Areas, Volumes, Weights, Displacement, etc. 43 

(1) Displacement in tons. 

(2) Area of immersed midship section. 

(3) Prismatic coefficient of displacement. 

Ans. (i) 7763 tons ; (2) 904 square feet ; (3) 0-675. 

41. A steam yacht is 144 feet long, 22' 6" broad, 9 feet draught ; dis- 
placement, 334 tons salt water j area of midship section, 124 square ieet. 
Find 

(1) Block coefficient of displacement. 

(2) Prismatic 

(3) Midship-section coefficient. 

Ans. (i) 0-4; (2) 0-655; (3) 0-612. 

42. Find the displacement in tons in salt water, area of the immersed 
midship section, prismatic coefficient of displacement, having given the 
following particulars : Length, i68feet ; breadth, 25 feet ; draught, lo'6"; 
midship-section coefficient, 0*87 ; block coefficient of displacement, 0-505. 

Ans. 750 tons ; 228-5 square feet ; 0*685. 

43. A vessel in the form of a box, 100 feet long, 10 feet broad, and 20 
feet deep, floats at a draught of 5 feet. Find the draught if a central 
compartment IO feet long is bilged below water. 

Ans. 5' 6J". 

44. In a given ship, pillars in the hold can be either solid iron 4! inches 
diameter, or hollow iron 6 inches diameter and half inch thick. Find the 
saving in weight for every 100 feet length of these pillars, if hollow pillars 
are adopted instead of solid, neglecting the effect of the solid heads and 
heels of the hollow pillars. 

Ans. i -35 ton. 

45. What is the solid contents of a tree whose girth (circumference) is 
60 inches, and length is 18 feet? 

Ans. 3 5 '8 cubic feet nearly. 

46. A portion of a cylindrical steel stern shaft casing is I2f feet long, 
ii inch thick, and its external diameter is 14 inches. Find its weight in 
pounds. 

Ans. 2170 Ibs. 

47. A floating body has a water-plane whose semi-ordinates 25 feet 
apart are 0-3, 8, 12, 10, 2 feet respectively, and every square station is in 
the form of a circle with its centre in the water-plane. Find the volume of 
displacement (TT = 3/&). 

Ans. 12,414 cubic feet. 

48. A quadrant of 16 feet radius is divided by means of ordinates parallel 
to one radius, and the following distances away : 4, 8, 10, 12, 13, 14, 15 
feet respectively. The lengths of these ordinates are found to be 15-49, 
13-86, 12-49, 10-58, 9-33, 7-75, and 5-57 feet respectively. Find 

(1) The exact area to two places of decimals. 

(2) The area by using only ordinates 4 feet apart. 

(3) The area by using also the half-ordinates. 

(4) The area by using all the ordinates given above. 

(5) The area as accurately as it is possible, supposing the ordinate 12-49 

had not been given. 
Ans. (i) 201-06; (2) 197*33; (3) 19975 5 (4) 200-59; (5) 200-50. 

49. A cylindrical vessel 50 feet long and 16 feet diameter floats at a 
constant draught of 12 feet in salt water. Using the information given in 
the previous question, find the displacement in tons. 

Ans. 231 tons nearly. 

50. A bunker 24 feet long has a mean section of the form of a trapezoid, 
with length of parallel sides 3 feet and 4*8 feet, and distance between them 
10'5 feet. Find the number of tons of coal contained in the bunker, assuming 



44 Theoretical Naval Architecture. 

I ton to occupy 43 cubic feet. If the parallel sides are perpendicular to 
one of the other sides, and the side 4*8 feet long is at the top of the section, 
where will the top of 17 tons of coal be, supposing it to be evenly 
distributed ? 

(This latter part should be done by a process of trial and error.) 

Ans. 22-8 tons ; 2' 3" below the top. 

51. The sections of a ship are 20 feet apart. A coal-bunker extends 
from 9 feet abaft No. 8 section to I foot abaft No. 15 section, the total 
length of the bunker thus being 132 feet. The areas of sections of the 
bunker at Nos. 8, II, and 15 are found to be 126, 177, and 145 square 
feet respectively. With this information given, estimate the capacity of the 
bunker, assuming 44 cubic feet of coal to go to the ton. Stations numbered 
from forward. 

Ans. 495 tons. 

52. The tons per inch immersion at water-lines 2 feet apart are 18-09, 
i6'8o, 15*15, I3'I5, I0 4 49, and 6-48. The draught of water to the top 
water-line is n' 6", and below the lowest water-line there is a displacement 
of 75 '3 tons. Find the displacement in tons, and construct a curve of 
displacement. 

Ans. 1712 tons. 

53. A tube 35 feet long, 16 feet diameter, closed at the ends, floats in 
salt water with its axis in the surface. Find approximately the thickness of 
the tube, supposed to be of iron, neglecting the weight of the ends. 

Ans. 0*27 foot. 

54. Find the floating power of a topmast, length 64 feet, mean diameter 
21 inches, the wood of the topmast weighing 36 Ibs. per cubic foot. 

(The floating power of a spar is the weight it will sustain, and this is 
the difference between its own weight and that of the water it displaces. 
In constructing a raft, it has to be borne in mind that all the weight of 
human beings is to be placed on it, and that a great quantity of provisions 
and water may be safely carried Hinder it. For instance, a cask of beef 
slung beneath would be 116 Ibs., above 300 Ibs. See " Sailor's Pocket- 
book," by Admiral Bedford.) 

Ans. 43 10 Ibs. 

55. Show that the following approximate values may be taken for the 
" tons per inch immersion " in salt water at the load draught : 

(1) For ships with fine ends ^ X L X B 

(2) ,, of ordinary form 555 X L X B 

(3) with bluff ends 5 fo X L X B 

L and B being the length and breadth respectively of the load water-plane. 

56. Show that a vessel passing from water of density d' into water^ of 

density d (<f being greater than d) will decrease her freeboard by 7p j- 

inches, where W is the displacement in tons and T the tons per inch 
immersion when in the denser water. 

A vessel 400 feet long, 45 feet broad, floats in Belfast water (ion ozs. 
to a cubic foot) at a draught of 21' 2^". By how much will the free- 
board be increased when in salt water (1025 ozs. to a cubic foot) ? (Coeffi- 
cient of fineness of displacement, 0*62 ; coefficient of fineness of L.W.P.,. 

07S-) 

Ans. 2 '9 inches. 



CHAPTER IL 

MOMENTS, CENTRE OF GRAVITY, CENTRE OF BUOY- 
ANCY, DISPLACEMENT TABLE, PLANIMETER, ETC. 

Principle of Moments. The moment of a force about 
any given line is the product of the force into the perpen- 
dicular distance of its line of action from that line. It may 
also be regarded as the tendency to turn about the line. A 
man pushes at the end of a capstan bar (as Fig. 27) with a 

w. 




FIG. 7 



certain force. The tendency of the capstan to turn about its 
axis is given by the force exerted by the man multiplied by 
his distance from the centre of the capstan, and this is the 
moment of the force about the axis. If P is the force exerted 
by the man in pounds (see Fig. 27), and d is his distance from 
the axis in feet, then 

The moment about the axis = P x d foot-lbs. 

The same moment can be obtained by a smaller force with 
a larger leverage, or a larger force with a smaller leverage, and 
the moment can be increased : 

(1) By increasing the force; 

(2) By increasing the distance of the force from the axis. 



4 6 



Theoretical Naval Architecture, 



If, in addition, there is another man helping the first man, 
exerting a force of F Ibs. at a distance from the axis of <? 
feet, the total moment about the axis is 



We must now distinguish between moments tending to turn 
one way and those tending to turn in the opposite direction. 

Thus, in the above case, we may take a rope being wound 
on to the drum of the capstan, hauling a weight W Ibs. If the 
radius of the drum be a feet, then the rope tends to turn the 
capstan in the opposite direction to the men, and the moment 
about the axis is given by 

W X a foot-lbs. 

If the weight is just balanced, then there is no tendency to 
turn, and hence no moment about the axis of the capstan, and 
leaving out of account all consideration of friction, we have 

(P X <*) + (F X O = W x a 

The most common forces we have to deal with are those 
caused by gravity, or the attraction of bodies to the earth. This 
is known as their weight, and the direction of these forces must 
gft be parallel at any given place. If we have a number of 
weights, Wi, W a , and W 8 , on a beam at A, B, and C (Fig. 28), 



n 



W. 



FIG. 28. 



whose end is fixed at O, the moment of these weights about O 
is given by 

(Wj X AO) + (W. X BO) + (W, X CO) 
This gives the tendency of the beam to turn about O, due to 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 47 

the weights W t , W 2 , and W 3 placed upon it, and the beam must 
be strong enough at O in order to resist this tendency, or, as 
it is termed, the bending moment. Now, we can evidently 
place a single weight W, equal to the sum of the weights 
W 1} W 2 , and W 3 , at some point on the beam so that its moment 
about O shall be the same as that due to the three weights. 
If P be this point, then we must have 

W x OP = (W t X OA) + (W 2 x OB) + (W 8 x OC) 
or, since W = W a + W 3 + W, 

OP = (W t X OA) + (W 2 x OB) + (W 8 x OC) 

Wl + W 2 + W, 

Example. Four weights, 30, 40, 50, 60 Ibs. respectively, are placed 
on a beam fixed at one end, O, at distances from O of 3, 4, 5, 6 feet 
respectively. Find the bending moment at O, and also the position of a 
single weight equal to the four weights which will give the same bending 
moment. 

Bending moment at O = (30 X 3) + (40 X 4) + (50 X 5) + (60 x 6) 
= 90 + 160 + 250 + 360 
= 860 foot-lbs. 
Total weight = 180 Ibs. 
.'. position of single weight = f($ = 4$ feet from O 

Centre of Gravity. The single weight W above, when 
placed at P, has the same effect on the beam at O as the 
three weights W x , W a , and W 8 . The point P is termed the 
centre of gravity of the weights W 1} W 2 , and W s . Thus we 
may define the centre of gravity of a number of weights as 
follows : 

The centre of gravity of a system of weights is that point at 
which we may regard the whole system as being concentrated. 

This definition will apply to the case of a solid body, since 
we may regard it as composed of a very large number of small 
particles, each of which has a definite weight and occupies 
a definite position. A homogeneous solid has the same 
density throughout its volume ; and all the solids with which 
we have to deal are taken as homogeneous unless otherwise 
specified. 

It follows, from the above definition of the centre of 
gravity, that if a body is suspended at its centre of gravity, 



48 Theoretical Naval Architecture. 

it would be perfectly balanced and have no tendency to move 
away from any position in which it might be placed. 

To Find the Position of the Centre of Gravity 
of a number of Weights lying in a Plane. Two lines 
are drawn in the plane at right angles, and the moment of the 
system of weights is found successively about each of these 
lines. The total weight being known, the distance of the 
centre of gravity from each of these lines is found, and conse- 
quently the position of the centre of gravity definitely fixed. 




FIG. 



The following example will illustrate the principles in- 
volved : Four weights, of 15, 3, 10, and 5 Ibs. respectively, 
are lying on a table in definite positions as shown in Fig. 29. 
Find the position of the centre of gravity of these weights. 
(If the legs of the table were removed, this would be the place 
where we should attach a rope to the table in order that it 
should remain horizontal, the weight of the table being 
neglected.) 



Moments, Centre of Gravity p , Centre of Buoyancy, etc. 49 

Draw two lines, Ox, Oy, at right angles on the table in 
any convenient position, and measure the distances of each of 
the weights from Ox, Oy respectively : these distances are 
indicated in the figure. The total weight is 33 Ibs. The 
moment of the weights about Ox is 

(i5 X 7) + (3 X 3) + (10 X 5) + (5 X 1-5) = 171-5 foot-lbs. 
The distance of the centre of gravity from Ox = = 5*2 feet 

If we draw a line A A a distance of 5*2 feet from Ox, the 
centre of gravity of the weights must be somewhere in the 
line AA. 

Similarly, we take moments about Oy, finding that the 
moment is 150 foot-lbs., and the distance of the centre of 
gravity from Oy is 

V# = 4*25 ^et 

If we draw a line BB a distance of 4*25 feet from Oy, the 
centre of gravity of the weights must be somewhere in the line 
BB. The point G, where AA and BB meet, will be the centre 
of gravity of the weights. 

Centres of Gravity of Plane Areas. A plane area has 
length and breadth, but no thickness, and in order to give a 
definite meaning to what is termed its centre of gravity, the 
area is supposed to be the surface of a thin lamina or plate of 
homogeneous material of uniform thickness. With this sup- 
position, the centre of gravity of a plane area is that point at 
which it can be suspended and remain in equilibrium. 

Centres of Gravity of Plane Figures. 

Circle. The centre of gravity of a circle is obviously at 
its centre. 

Square and Rectangle. The centre of gravity of 
either of these figures is at the point where the diagonals 
intersect. 

Rhombus and Rhomboid. The centre of gravity of 
either of these figures is at the point where the diagonals 
intersect. 



Theoretical Naval Architecture. 




D. 

FIG. 30. 



Triangle. Take the triangle ABC, Fig. 30. Bisect any 
two sides BC, AC in the points D and E. Join AD, BE. The 
point G where these two lines intersect is the centre of gravity 

of the triangle. It can be 
proved that the point G is 
situated so that DG is one-third 
DA, and EG is one-third EB. 
We therefore have the following 
rules : 

i. Bisect any two sides of 
the triangle, and join the points 
thus obtained to the opposite angu- 
lar points. Then the point in 
which these two lines intersect is the centre of gravity of the triangle. 
2. Bisect any side of the triangle, and join the point thus 
obtained with the opposite angular point. The centre of gravity 
of the triangle will be on this line, and at a point at one-third its 
length measured from the bisected side. 

Trapezium. Let ABCD, Fig. 31, be a trapezium. By 
joining the corners A and C we can divide the figure into two 
triangles, ADC, ABC. The centres of gravity, E and F, of 

these triangles can be 
found as indicated 
above. Join EF. The 
centre of gravity of the 
whole figure must be 
somewhere in the line 
EF. Again, join the 
corners D and B, thus 
dividing the figure into 
two triangles ADB, 
CDB. The centres of 
triangles can be found. The 




D. C 

FIG. 31. 

gravity, H and K, of these 
centre of gravity of the whole figure must be somewhere in the 
line HK ; therefore the point G, where the lines HK and EF 
intersect, must be the centre of gravity of the trapezium. 

The following is a more convenient method of finding the 
centre of gravity of a trapezium. 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 5 l 

Let ABCD, Fig. 32, be a trapezium. Draw the diagonals 
AC, BD, intersecting at E. In the figure CE is greater than 



a 




FIG. 39. 

EA, and DE is greater than EB. Make CH = EA and DF 
= EB. Join FH. Then the centre of gravity of the triangle 
EFH will also be the centre of gravity of the trapezium ABCD. 1 

(A useful exercise in drawing would be to take a trapezium 
on a large scale and find its 
centre of gravity by each of 
the above methods. If the 
drawing is accurately done, the 
point should be in precisely 
the same position as found by 
each method.) 

To find the Centre of 
Gravity of a Plane Area 
by Experiment. Draw out 
the area on a piece of card- 
board or stiff paper, and cut 
out the shape. Then suspend 
the cardboard as indicated in 
Fig- 33, a small weight, W, 
being allowed to hang plumb. 
A line drawn behind the string AW must pass through the 
centre of gravity. Mark on the cardboard two points on the 
string, as A and B, and join. Then the centre of gravity must 
lie on AB. Now suspend the cardboard by another point, C, 
1 See Example 22, for C.G. of a trapezoid. 




Theoretical Naval Architecture. 



as in Fig. 34, and draw the line CD immediately behind the 
string of the plumb-bob W. Then also the centre of gravity 
must lie on the line CD. Consequently it follows that the 

point of intersection G of the 
lines AB and CD must be the 
centre of gravity of the given 
area. 

Example. Set out the section of 
a beam on a piece of stiff paper, and 
find by experiment the position of its 
centre of gravity, the beam being formed 
of a bulb plate 9 inches deep and 
i inch thick, having two angles on the 
upper edge, each 3" x 3" X ". 

Ans. 3 inches from the top. 




Centres of Gravity of Solids 
formed of Homogeneous 
Material. 

Sphere. The centre of 
gravity of a sphere is at its centre. 
Cylinder. The centre of 
gravity of a cylinder is at one- 
half its height from the base, on 
the line joining the centres of 
gravity of the ends. 
Pyramid or Cone. The centre of gravity of a pyramid 
or cone is at one- fourth the height of the apex from the base, 
on the line joining the centre of gravity of the base to the 
apex. 

Moment of an Area. 

The geometrical moment of a plane area relatively to a 
given axis, is the product of its area into the perpendicular 
distance of its centre of gravity from the given axis. It follows 
that the position of the centre of gravity is known relatively to 
the given axis if we know the geometrical moment about the 
axis and also the area, for the distance will be the moment 
divided by the area. It is usnal to speak of the moment of an 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 53 

area about a given axis when the geometrical moment is really 
meant. 

To find the Position of the Centre of Gravity of 
a Curvilinear Area with respect to one of its Ordi- 
nates. Let AEDO, Fig. 35, be a plane curvilinear area, and 




we wish to find its centre of gravity with respect to the end 
ordinate, OA. To do this, we must first find the moment of 
the total area about OA, and this divided by the area of the 
figure itself will give the distance of the centre of gravity from 
OA. Take any ordinate, PQ, a distance of x from OA, and 
at PQ draw a strip A* wide. Then the area of the strip is 
y x A# very nearly, and the moment of the strip about OA is 
(y x &x)x very nearly. 

If now A* be made indefinitely small, the moment of the 
strip about OA will be 

y . x . dx 

Now, we can imagine the whole area divided up into such 
strips, and if we added up the moments about OA of all such 
strips, we should obtain the total moment about OA. Therefore, 
using the notation we employed for finding the area of a plane 
curvilinear figure on p. 14, we shall have 

Moment of the total area about OA = jy . x . dx 
The expression for the area i 



54 



Theoretical Naval Architecture. 



and this is of the same form as the expression for the moment. 
Therefore, instead of y we put yx through Simpson's rule in 
the ordinary way, and the result will be the moment about OA. 
Set off on BC a length BF = BC X h, and on DE a length 
DG = DE x 2h. Then draw through all such points a curve, 
as OFG. 1 Any ordinate of this curve will give the ordinate of 
the original curve at that point multiplied by its distance from 
OA. The area of an elementary strip of this new curve will be 
y . x . dx, and the total area of the new curve will be jy . x . dx, 
or the moment of the original figure about OA. Therefore, to 
find the moment of a curvilinear figure about an end ordinate, 
we take each ordinate and multiply it by its distance from the 
end ordinate. These products, put through Simpson's rule in 
the ordinary way, will give the moment of the figure about the 
end ordinate. This moment divided by the area will give the 
distance of the centre of gravity of the area from the end 
ordinate. 

Example. A midship section has semi-ordinates, i' 6" apart, com- 
mencing at the L.W.L., of length 8'6o, 8'io, 6-95, 4-90, 275, 1-50, 070 
feet respectively. Find the area of the section and the distance of its C.G. 
from the L.W.L, 



Number of 
ordinates. 


Length of 
ordinates. 


Simpson's 
multipliers. 


Function of 
ordinates. 


Number of 
intervals from 
No. i ord. 


Products for 
moment. 


, 


8'60 


, 


8'60 





O'O 


2 


8-10 


4 


32-40 


I 


32-40 


3 


6'95 


2 


I3-00 


2 


27-80 


4 


4-90 


4 


19-60 


3 


58-80 


j 


275 


2 


S'SO 


4 


22-00 


6 


1-50 


4 


6"oo 


5 


30-00 


7 


070 


1 


0-70 


6 


4-20 



86-70 



Function of 
moment 



;-2o 



The half-area will be given by 8670 X (3 X 1*5) = 43'35 square feet 
and the whole area is 86*70 

The arrangement above is adopted in order to save labour in 

1 If the original curve is a parabola of the second order, this curve will 
be one of the third order, and it can be proved that Simpson's first rule 
will integrate a parabola of the third order. 



Moments ', Centre of Gravity, Centre of Buoyancy, etc. 55 

finding the moment of the area. In the fourth column we have 
the functions of the ordinates, or the ordinates multiplied succes- 
sively by their proper multipliers. In the fifth column is placed, 
not the actual distance of each ordinate from the No. i ordi- 
nate, but the number of intervals away, and the distance apart 
is brought in at the end. In the sixth column the products of 
the functions in column 4 and the multipliers in column 5 are 
placed. It will be noticed that we have put the ordinates 
through Simpson's multipliers first, and then multiplied by the 
numbers in the fifth column after. This is the reverse to the 
rule given in words above, which was put into that form in 
order to bring out the principle involved more plainly. The 
final result will, of course, be the same in either case, the 
method adopted giving the result with the least amount of 
labour, because column 4 is wanted for finding the area. The 
sum of the products in column 6 will not be the moment 
required, because it has to be multiplied as follows : First, by 
one-third the common interval, and second, by the distance 
apart of the ordinates. 

The moment of the half-area ) 

about the L.W.L. I = W 2 X X '*> X * 



and the distance of the C.G. of the half-area from the L.W.L. 
is 

1 31*4 

Moment -r area =* - - = 3-03 feet 
43'35 

It will be noticed that we have multiplied both columns 
4 and 6 by one-third the common interval, the distance of the 
C.G. from No. i ordinate being obtained by 

175-20 X (JX 1-5) X 1-5 
86-70 X ft X 1-5) 

The expression ^ X 1*5 is common to both top and bottom, 
and so can be cancelled out, and we have 

175-20 x i'5 

^7 * = 3-03 feet 



56 Theoretical Naval Architecture. 

The position of the centre of gravity of the half-area with 
regard to the L.W.L. is evidently the same as that of the whole 
area. 

When finding the centre of gravity of a large area, such as 
a water-plane of a vessel, it is usual to take moments about 
the middle ordinate. This considerably simplifies the work, 
because the multipliers in column 5 are not so large. 

Example. T^e semi-ordinates of the load water-plane of a vessel 395 
feet long are, commencing from forward, o, 10*2, 20*0, 27*4, 32*1, 34*0", 
33'8, 31-7, 27-6, 20-6, 9-4. Find the area and the distance of its C.G. 
from the middle ordinate. 

In addition to the above, there is an appendage abaft the last ordinate, 
having an area of 153 square feet, and whose C.G. is 5 '6 feet abaft the last 
ordinate. Taking this appendage into account, find the area and the 
position of the C.G. of the water-plane. 



Number of 
ordinates. 


Length of 
ordinates. 


Simpson's 
multipliers. 


Function of 
ordinal.es. 


Number of 
interval from 
mid. ord. 


Product for 
moment. 


, 


O'O 


I 


0'0 


5 


O'O 


2 


IO"2 


4 


40-8 


4 


163-2 


3 


20'0 


2 


40 - o 


3 


I2O'O 


4 


27 '4 


4 


io9'6 


2 


2I9-2 


S 


3 2 I 


2 


64-2 


I 


6 4 -2 


6 


34'0 


4 


136-0 


O 


566-6 


7 


33-8 


2 


67-6 


I 


67-6 


8 


317 


4 


126-8 


2 


253*6 


9 


27-6 


2 


55 '2 


3 


165 6 


10 


20 '6 


4 


82-4 


4 


329-6 


ii 


9'4 


I 


9'4 


5 


47-0 



732-0 863-4 

The half-area will be given by 

732-0 X (J X 39-5; = 9638 square feet 

The fifth column gives the number of intervals away from the middle 
ordinate, and the products are obtained for the forward portion adding up 
to 566*6, and they are obtained for the after portion adding up to 863*4. 
This gives an excess aft of 863-4 566-6 = 296*8. The distance of the C.G, 
abaft the middle ordinate is then given by 



The area of both sides is 19,276 square feet. 

The second part of the question takes into account an appendage abaft 
No. 1 1 ordinate, having an area of 153 square feet. 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 57 

The total area will then be 

19,276 + 153 = 19,429 square feet 

To find the position of the C G. of the whole water-plane, we take 
moments about ND. 6 ordinate, the distance of the C.G. of the appendage 
from it being 

1 97'S + 5' 6 = 2 3'i feet 

Moment of main area abaft No. ordinate = 19,276 X i6'oi = 308,609 
appendage = 153 X 203-1 = 31,074 

/.total moment abaft No. 6 ordinate = 308,609 -f- 31,074 



and the distance of the centre of gravity \ 339683 _ . 
of the whole area abaft No. 6 ordinate / = I9429 ~ r 7'4 i 

To find the Position of the Centre of Gravity of 
a Curvilinear Area contained between Two Con- 
secutive Ordinates with respect to the Near End 
Ordinate. The rule investigated in the previous paragraph 
for finding the centre of gravity of an area about its end ordi- 
nate fails when applied to such a case as the above. For 
instance, try the following example : 

A curve has ordinates 10, 9, 7 feet long, 4 feet apart. To 
find the position of the centre of gravity of the portion between 
the two first ordinates with respect to the end ordinate. 



Ordinates. 


Simpson's 
multipliers. 


Functions. 


Multipliers 
for moment. 


Products for 
moment. 


IO 


5 


5 





O 


9 


8 


72 


I 


72 


7 


i 


-7 


2 


-I 4 



. us 5 

Centre of gravity from the end ordinate would be 
58 X 4 < 



Now this is evidently wrong, since the shape of the curve is 
such that the centre of gravity ought to be slightly less than 
2 feet from the end ordinate. 

We must use the following rule : 

To ten times the middle ordinate add three times the near 
end ordinate and subtract the far end ordinate. Multiply tht 






Theoretical Naval Architecture. 



remainder by one-twenty-fourth the square of the common interval. 
The product will be (he moment about the end ordinate. 

Using jj, j/ 2 , y^ for the lengths of the ordinates, and h the 
common interval, the moment of the portion between the 
ordinates y and y 2 about the ordinate y l is given by 



We will now apply this rule to the case considered above. 



OrHinafPC 


Area. 


Moment. 




Simpson's 
multipliers. 


Functions. 


Simpson's 
multipliers. 


Functions. 


to 
9 


I 


50 
72 


3 

10 


30 

9 


7 


1 


7 


i 


- 7 



Moment 
Area 



"3 



Therefore distance of the centre of gravity from the end oidi- 
nate is 

"3 X H = IJ 3 X 2 x 3 
115 X Yt 115 X 3 

= fff =I '965 feet 

This result is what one might expect by considering the 
shape of the curve. 

To find the Position of the Centre of Gravity of a 
Curvilinear Area with respect to its Base. Let DABC, 
Fig- 36, be a plane curvilinear area. We wish to find the 
distance of its centre of gravity from the base DC. To do 
this, we must first find the moment of the figure about DC and 
divide it by the area. Take any ordinate PQ, and at PQ 
draw a consecutive ordinate giving a strip A# wide. Then the 
area of the strip is 

y x A.* very nearly 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 59 



and regarding it as a rectangle, its centre of gravity is at a 
distance of \y from the base. Therefore the moment of the 
strip about the base is Q 

\f X A* 

If now we consider the strip 
to be indefinitely thin, its 
moment about the base will 
be 



and the moment of the total D 

area about the base must be 

the sum of the moments of all such strips, 01 



p 



FIG. 36. 



This expression for the moment is of the same form as that for 
the area, viz. jy . dx. Therefore, instead of y we put \y* l 
through Simpson's rule in the ordinary way, and the result will 
be the moment of the area about DC. 

Example. An athwartship coal-bunker is 6 feet long in a fore-and-aft 
direction. It is bounded at the sides by two longitudinal bulkheads 34 feet 
apart, and by a horizontal line at the top. The bottom is formed by the 
inner bottom of the ship, and is in the form of a curve having vertical 
ordinates measured from the top of 12*5, 15*0, 16*0, l6'3, 16*4, 16*3, 16*0, 
15*0, 1 2* 5 feet respectively, the first and last ordinates being on the bulk- 
heads. Find 

(1) The number of tons of coal the bunker will hold. 

(2) The distance of the centre of gravity of the coal from the top. 
The inner bottom is symmetrical either side of the middle line, so we 

need only deal with one side. The work is arranged as follows ; 



Ordinates. 


Simpson's 
multipliers. 


Functions of 
ordinates. 


Squares of 
ordinates. 


Simpson's 
multipliers. 


Functions 
of squares. 


I6'4 


I 


16-4 


269 


I 


269 


l6- 3 


4 


6 5 -2 


266 


4 


1064 


16-0 


2 


3 2-0 


2 5 6 


2 


512 


15-0 


4 


OO'O 


225 


4 


900 


12-5 


I 


12-5 


I 5 6 


i 


I 5 6 



Function of area 186*1 



Function of moment 2901 



1 This assumes that Simpson's first rule, which will most probably be 
used, will correctly integrate a parabola of the fourth order, which can be 
shown to be the case for all practical purposes. 



60 Theoretical Naval Architecture. 

Common interval = 4*25 feet 
Half-area of section = l86'i X J X 4*25 square feet 

Volume of bunker = iS6'i x 4 ' 2g x 2 x 6 cub i c feet 

Number of tons of coal = 186*1 X |J 
= 72 tons 

Moment of half-area below top = 2901 X - X ^^ 
And distance of C.G. from the top = - 



_ 



= 78 feet. 

In the first three columns we proceed in the ordinary way 
for finding the area. In the fourth column is placed, not the 
half-squares, but the squares of the ordinates in column i, the 
multiplication by \ being brought in at the end. These 
squares are then put through Simpson's multipliers, and the 
addition of column 6 will give a function of the moment of 
the area about the base. This multiplied by \ and by \ the 
common interval gives the actual moment This moment 
divided by the area gives the distance of the centre of gravity 
we want. It will be noticed that \ the common interval 
comes in top and bottom, so that we divide the function of the 
moment 2901 by the function of the area 186*1, and then 
multiply by \ to get the distance of centre of gravity required. 

It is not often required in practice to find the centre of 
gravity of an area with respect to its base, because most of the 
areas we have to deal with are symmetrical either side of a 
centre line (as water-planes), but the problem sometimes occurs, 
the question above being an example. 

To find the Position of the Centre of Gravity of 
an Area bounded by a Curve and Two Radii. We 
have already seen (p. 15) how to find the area of a figure such 
as this. It is simply a step further to find the position of the 
centre of gravity with reference to either of the bounding radii. 
Let OAB, Fig. 13, be a figure bounded by a curve, AB, and 
two bounding radii, OA, OB. Take any radius OP, the angle 
BOP being called * 0, and the length of OP being called r. 



Moments, Centre of Gravity, Centre of Buoyancy,, etc. 6 1 

Draw a consecutive radius, OP' ; the angle POP' being indefi- 
nitely small, we may call it dd. Using the assumptions we 
have already employed in finding areas, the area POP' = 
Jr 2 .<#?, POP' being regarded as a triangle. The centre of 
gravity of POP' is at g t and Og = fr, and gm is drawn perpen- 
dicular to OB, andgm = \r . sin 6 (see p. 91). 



- *>>< **<> 

= r*.sin Q.dB 

The moment of the whole figure about OB is the sum of 
the moments of all such small areas as POP', or, using the 
ordinary notation 



This is precisely similar in form to the expression we found 
for the area of such a figure as the above (see p. 15), viz. 



so that, instead of putting ^ through Simpson's rule, measuring 
r at equidistant angular intervals, we put \r* . sin 6 through 
the rule in a similar way. This will be best illustrated by the 
following example : 

Example. Find the area and position of centre of gravity of a quadrant 
of a circle with reference to one of its bounding radii, the radius being 
10 feet. 

We will divide the quadrant by radii 15 apart, and thus be able to use 
Simpson's first rule. 



O . 




. 


,'d 





. 


Ij 




tt" 


* d 


-R , 


Number 
radius 


fi 


a 

I 


c .Ji 

if 


Product 
area. 


I 


H 

if 

<< 1C 


If 


l.fl 

Ix 

*\ 


.a 

fl 


Product 
momen 


I 


10 


IOO 


I 


IOO 


IOOO 





O'O 


O 


i 


O 


2 


IO 


IOO 


4 


4OO 


IOOO 


15 


0-258 


258 


4 


1,032 


3 


10 


IOO 


2 


200 


IOOO 


30 


0-500 


500 


2 


I,OOO 


4 


10 


IOO 


4 


4OO 


IOOO 


45 


0-707 


707 


4 


2,828 




10 
IO 


IOO 
IOO 


2 

4 


200 

4 00 


IOOO 
IOOO 


60 

75 


0-866 
0-965 


866 
965 


2 

4 


1,732 

3,860 


7 


10 


IOO 


I 


IOO 


IOOO 


90 


I '000 


IOOO 


I 


1,000 



Function of area 1800 



Function of moment 11,452 



62 Theoretical Naval Architecture. 

The circular measure of 180 = IT = 3-1416 




= ?8'54 square feet 
Moment of area about the first radius = 11,452 X -X f -x - J 

therefore distance of centre of gravity from the first radius is 



Moment -f- area = 



11452 x 2 
= l8cx>x 3 = 4-24feet 



The exact distance of the centre of gravity of a quadrant 

from either of its bounding radii is times the radius, and if 

3 71 " 

this is applied to the above example, it will be found that the 
result is correct to two places of decimals, and would have 
been more correct if we had put in the values of the sines of 
the angles to a larger number of decimal places. 

Centre of Gravity of a Solid Body which is 
bounded by a Curved Surface and a Plane. In the 
first chapter we saw that the finding the volume of such a solid 
as this was similar in principle to the finding the area of a 
plane curve, the only difference being that we substitute areas 
for simple ordinates, and as a result get the volume required. 
The operation of finding the centre of gravity of a volume in 
relation to one of the dividing planes is precisely similar to the 
operation of finding the centre of gravity of a curvilinear area 
in relation to one of its ordinates. This will be illustrated by 
the following example : 

Example. A coal-bunker has sections 17* 6" apart, and the areas of 
these sections, commencing from forward, are 98, 123, 137, 135, 122 square 
feet respectively. Find the volume of the bunker, and the position of its 
centre of gravity in a fore-and-aft direction. 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 63 



Areas. 


Simpson's 
multipliers. 


Functions of 
areas. 


Number of 
intervals from 
forward. 


Products for 
moments. 


9 8 


I 


9 8 





o 


123 


4 


492 


I 


492 


137 


2 


274 


2 


548 


135 


4 


540 


3 


l62O 


122 


I 


122 


4 


4 88 



1526 



3148 



Volume =1526x^x17$ = 8902 cubic feet 
moment = 3148 x \ X 17$ X 17$ 
/. distance of centre of gravity \ _ 3*48 X I7'5 
from forward end / 1526 



36*2 feet 



It is always advisable to roughly check any result such 
as this ; and if this habit is formed, it will often prevent mis- 
takes being made. The total length of this bunker is 4 x 
17' 6" = 70 feet, and the areas of the sections show that the 
bunker is fuller aft than forward, and so, on the face of it, we 
should expect the position of the centre of gravity to be some- 
what abaft the middle of the length ; and this is shown to be 
so by the result of the calculation. Also as regards the volume. 
This must be less than the volume of a solid 70 feet long, and 
having a constant section equal to the area of the middle 
section of the bunker. The volume of such a body would 
be 70 X 137 = 9590 cubic feet. The volume, as found by 
the calculation, is 8902 cubic feet, thus giving a coefficient of 
IHHrf = '93 nearly, which is a reasonable result to expect. 

Centre of Buoyancy. The centre of buoyancy of a 
vessel is the centre of gravity of the underwater volume, or, 
more simply, the centre of gravity of the displaced water. 
This has nothing whatever to do with the centre of gravity of 
the ship herself. The centre of buoyancy is determined solely 
by the shape of the underwater portion of the ship. The 
centre of gravity of the ship is determined by the distribution 
of the weights forming the structure, and of all the weights she 
has on board. Take the case of two sister ships built from 
the same lines, and each carrying the same weight of cargo 
and floating at the same water-line. The centre of buoyancy 



6 4 



Theoretical Naval Architecttire. 



of each of these ships must necessarily be in the same position. 
But suppose they are engaged in different trades the first, 
say, carrying a cargo of steel rails and other heavy weights, 
which are stowed low down. The second, we may suppose, 
carries a cargo of homogeneous materials, and this has to be 
stowed much higher than the cargo in the first vessel. It is 
evident that the centre of gravity in the first vessel must be 
much lower down than in the second, although as regards form 
they are precisely similar. This distinction between the centre 
of buoyancy and the centre of gravity is a very important 
one, and should always be borne in mind. 

To find the Position of the Centre of Buoyancy of 
a Vessel in a Fore-and-aft Direction, having given 
the Areas of Equidistant Transverse Sections. The 
following example will illustrate the principles involved : 

Example. The underwater portion of a vessel is divided by transverse 
sections 10 feet apart of the following areas, commencing from forward : 0*2, 
227, 48-8, 73-2, 88-4, 82-8, 587, 26-2, 3-9 square feet respectively. Find 
the position of the centre of buoyancy relative to the middle section. 



Number of 
station. 


Area of 

section. 


Simpson's 
multipliers. 


Functions of 
area. 


Number of 
intervals 
from middle. 


Product for 
moment. 


, 


0'2 


, 


0'2 


4 


0-8 


2 


22'7 


4 


90-8 


3 


272-4 


3 


48-8 


2 


97'6 


a 


195-2 


4 


73 -2 


4 


292-8 


i 


292-8 


5 


88'4 


2 


1768 





761-2 


6 


82-8 


4 


331'2 


i 


33i'2 


7 


587 


2 


II7'4 


2 


234-8 


8 


26-2 


4 


I04-8 


3 


3i4'4 


9 


3'9 




3'9 


4 


15-6 



Function of displacement 1215-5 



Function of 
moment 



\ 8 9 6-c 



Volume of displacement = 1215-5 X g 

excess of products aft = 896*0 761*2 = 134*8 
moment aft = 134-8 X ^ X 10 

C.B. abaft middle = '34* x x 10 

I2I5-5 * 5 Q 



1215-5 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 65 

The centre of gravity of a plane area is fully determined 
when we know its position relative to two lines in the plane, 
which are generally taken at right angles to one another. The 
centre of gravity of a volume is fully determined when we 
know its position relative to three planes, which are generally 
taken at right angles to one another. In the case of the under- 
water volume of a ship, we need only calculate the position of 
its centre of gravity relative to (i) the load water-plane, and 
(2) an athwartship section (usually the section amidships), 
because, the two sides of the ship being identical, the centre 
of gravity of the displacement must lie in the middle-line 
longitudinal plane of the ship. 

Approximate Position of the Centre of Buoyancy. 
In vessels of ordinary form, it is found that the distance of the 
centre of buoyancy below the L.W.L. varies from about -^ to -^ 
of the mean draught to top of keel, the latter being the case 
in vessels of full form. For yachts and vessels of unusual 
form, such a rule as this cannot be employed. 

Example, A vessel 13' 3" mean draught has her C.B. 5*34 feet below 
L.^V.L. 

Here the proportion of the draught is 

5 '34 8 -06 

.sLsCL = o'4.O1 = 

13-25 4 6 20 

This is an example of a fine vessel. 

Example. A vessel 27' 6" mean draught has her C.B. 12*02 feet below 
L.W.L. 

Here the proportion of the draught is 

12-02 __ 875 
27-5 ~ 20 

This is an example of a fuller vessel than the first case. 
Morrish's Approximate Formula for the Distance 
of the Centre of Buoyancy below the Load Water- 
line. 1 

Let V = volume of displacement up to the load-line in 

cubic feet ; 

A = the area of the load water-plane in square feet ; 
d the mean draught (to top of keel) in feet. 

1 See a paper in Transactions of the Institution of Naval Architects, by 
Mr. S. W. F. Morrish. M.I.N.A., in 1892. 



66 



Theoretical Naval Architecture. 



Then centre of buoyancy below L.W.L. = J ( - + % 



This rule gives exceedingly good results for vessels of 
ordinary form. In the early stages of a design the above 
particulars would be known as some of the elements of the 
design, and so the vertical position of the centre of buoyancy 
can be located very nearly indeed. In cases in which the 
stability of the vessel has to be approximated to, it is important 
to know where the C.B. is, as will be seen later when we are 
dealing with the question of stability. 

The rule is based upon a very ingenious assumption, as 
follows : 

In Fig. 36A, let ABC be the curve of water- plane areas, DC 
being the mean draught d. Draw the rectangle AFCD. Make 

y 
DE = -r D say. Draw EG parallel to DA cutting the diagonal 

*TL 

FD in H. Finish the figure as indicated. Then the assumption 
A. _ D. 





E. 



C. 



FIG. 36A. 



made is that the C.G. of the area DAHC is the same distance below 
the water-line as the C.G. of DABC which latter, of course, gives 
the distance below the water-line of the centre of buoyancy. It is 
seen by inspection that the assumption is a reasonable one. 

DAHC and DABC have the same area as we now proceed to 
show. The rectangles AH and HC are equal, so that the triangles 
AGH and HEC are equal, and therefore 

Area of AHCD = area of AGED 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 67 

The latter gives the volume of displacement, as it is a rectangle 
having sides equal to A and -r respectively. The area of DABC 

also gives the volume of displacement, so that DAHC and DABC 
are equal. 

We now have to determine the distance of the C.G. of DAHC 
below the water-line. 

Area AGH \ X AG X GH GH 

Area AGED ~ AG X AD & ' AD 



d being the draught. 

/. Area AGH = \- * rectangle AGED 



We may regard the figure DAHC made up by taking away 
AGH from the rectangle AE and putting it in the position HEC. 
The shift of its C.G. downwards is \.d. Therefore the C.G. of 
the whole figure will shift downwards, using the principle explained 
in p. 100, a distance x, given by 

AGED x x = AGH x - 
or putting in the value found above for the area of AGH, we have 



The C.G. of AGED is a distance below the water-line. There- 

2 

.fore the C.G. of DAHC is below the water-line, a distance 



which is Morrish's approximation to the distance of the C.B. below 
the water-line. 

The Area of a Curve of Displacement divided by 
the Load Displacement gives the Distance of the 
Centre of Buoyancy below the Load Water-line. This 
is an interesting property of the curve of displacement. The 
demonstration is as follows : 

Let OBL, Fig. 363, be the curve of displacement of a vessel 
constructed in the ordinary way, OW being the mean draught and 
WL being the displacement in tons. 

Take two level lines AB, A'B', a short distance apart, &z say. 
Call the area of the water-plane at the level of AB, A square feet, 



68 



Theoretical Naval Architecture. 



and the distance of this water-plane below the WL, z. The 
volume between AB and A'B' is A X As-, or supposing they are 
indefinitely close together A X dz. The moment of this layer 
about the WL is A x dz x (z + \ . dz) = A . z . dz, neglecting the 
t C'C. 



or 



FIG. 368. 

square of the small quantity dz. The distance of the C.B. below 
WL is the sum of all such moments divided by the displacement 
volume, 

j&.z.dz 

35 x WL 

Now the difference between the lengths of A'B' and AB is the 
weight of the water between these level lines or -% . A . dz. The 
area of the whole figure is given by the summatim of all such 
areas as the strip B'C, which has a length z and a breadth 
^g- . Adk. Area of figure is therefore aV/A- .z.dz, and this divided 
by the displacement is 



35 x WL 

which is the expression found above for the distance of the C.B. 
below the WL. 

Example. Draw a curve of displacement for all draughts of a cylindrical 
vessel 20 feet diameter and 150 feet long, and find by using the curve the 
distance of the C.B. from the base when floating (a) at 10 feet level draught, 
(b) at 15 feet level draught. 

Ans. (a) 576 feet ; (b) 8-25 feet. 

If a new curve be drawn having for ordinates the area of the curve of 
displacement at respective levels, it may readily be shown that the tangent 
to this curve at any draught will intersect the scale of draughts at the 
height of the centre of buoyancy. This new curve is a curve of moments 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 69 

of displacement up to each level line about such level line. By construct- 
ing such a curve in the graphic method of finding displacement (see later), 
considerable simplification of the process is obtained. Thus, in Fig. 39, 
AHL is the curve of displacement. By integrating this curve (still by the 
graphic method), a curve of moments of displacements is obtained, the 
ordinate of which on BR will be the moment of displacement BL about 
the L.W.L. This moment, divided by the displacement BL, gives the 
distance of the C.B. below the L.W.L. This may be checked by drawing 
the tangent to the new curve, as seen above. In a similar manner, in 
Fig. 40 the difference between the areas of BB in the fore and after bodies 
divided by the total displacement gives the fore and aft position of the C.B. 
with reference to No. 6 station. 

Displacement Sheet. 1 We now proceed to investigate 
the method that is very generally employed in practice to find 
the displacement of a vessel, and also the position of its centre 
of buoyancy both in a longitudinal and a vertical direction. 
The calculation is performed on what is termed a " Displace- 
ment Sheet" or "Displacement Table" and a specimen calcula- 
tion is given at the end of the book for a single-screw tug of 
the following dimensions : 

Length between perpendiculars 74' o' 

Breadth moulded 14' 6' 

Depth moulded 8' 3' 

Draught moulded forward 5' 5' 

aft 6' 2' 

,, mean 5' 9 

The sheer drawing of the vessel is given on Plate I. This 
drawing consists of three portions the body plan, the half- 
breadth plan, and the sheer. The sheer plan shows the ship 
in side elevation, the load water-line being horizontal, and the 
keel, in this case, sloping down from forward to aft. The ship 
is supposed cut by a number of transverse vertical planes, 
which are shown in the sheer plan as straight lines, numbered 
i, 2, 3, etc. Now, each of these transverse sections of the ship 
has a definite shape, and the form of each half-section to the 
outside of frames is shown in the body-plan, the sections being 
numbered as in the sheer. The sections of the forward end 
form what is termed the "fore-body? and those of the after 
end the " after-body" Again, the ship may be supposed to be 
cut by a series of equidistant horizontal planes, of which the 

1 For displacement sheet with combination of Simpson's first rule and 
Tchebycheffs rule, see Appendix A. 



70 Theoretical Naval Architecture. 

load water-plane is one. The shape of the curve traced on each 
of these planes by the moulded surface of the ship is given in 
the half-breadth plan, and the curves are numbered A, i, 2, 3, 
etc., to agree with the corresponding lines in the sheer and 
body plan. Each of these plans must agree with the other 
two. Take a special station, for example, No. 4. The breadth 
of the ship at No. 4 station at the level of No. 3 water-plane is 
Oa' in the body-plan, but it is also given in the half-breadth plan 
by Oa, and therefore Oa must exactly equal Oa'. The process 
of making all such points correspond exactly is known as 
"fairing? For full information as to the methods adopted in 
fairing, the student is referred to the works on " Laying-off" 
given below. 1 For purposes of reference, the dimensions of 
the vessel and other particulars are placed at the top of the 
displacement sheet. The water-lines are arranged on the 
sheer drawing with a view to this calculation, and in this case 
are spaced at an equidistant spacing apart of i foot, with an 
intermediate water-line between Nos. 5 and 6. The number 
of water-lines is such that Simpson's first rule can be used, and 
the multipliers are, commencing with the load water-plane 

i 4 2 4 ji 2 | 

The close spacing near the bottom is very necessary to 
ensure accuracy, as the curvature of sections amidships of the 
vessel is very sharp as the bottom is approached, and, as we 
saw on p. 13, Simpson's rules cannot accurately deal with areas 
such as these unless intermediate ordinates are introduced. 
Below No. 6 water-plane there is a volume the depth of which 
increases as we go aft, and the sections of this volume are very 
nearly triangles. This volume is dealt with separately on the 
left-hand side of the table, and is termed an " appendage? 

In order to find the volume of displacement between water- 
planes i and 6, we can first determine the areas of the water- 
planes, and then put these areas through Simpson's rule. To 
find the area of any of the water-planes, we must proceed in 
the ordinary manner and divide its length by ordinates so that 

1 " Laying Off," by Mr. S. J. P. Thearle ; " Laying Off," by Mr. T. II. 
Watson ; " Laying Off," by Messrs. Alt wood and Cooper. 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 7 1 

Simpson's rule (preferably the first rule) can be used. In the 
case before us, the length is from the after-edge of the stem to 
the forward edge of the body post, viz. 71 feet, and this 
length is divided into ten equal parts, giving ordinates to each 
of the water-planes at a distance apart of 7*1 feet. The dis- 
placement-sheet is arranged so that we can put the lengths of 
the semi-ordinates of the water-planes in the columns headed 
respectively L.W.L., 2\Y.L., 3"W.L., etc., the semi-ordinates at 
the several stations being placed in the same line as the numbers 
of ordinates given at the extreme left of the table. The lengths 
of the semi-ordinates are shown in italics. Thus, for instance, 
the lengths of the semi-ordinates of No. 3 W.L., as measured 
off, are 0-05, 1*82, 4*05, 5-90, 6-90, 7-25, 7-04, 6-51 5^ 
2*85, and 0*05 feet, commencing with the forward ordinate 
No. i, and these are put down in italics 1 as shown beneath 
the heading 3 W.L. in the table. The columns under the 
heading of each W.L. are divided into two, the semi-ordinates 
being placed in the first column. In the second column of each 
water-line is placed the product obtained by multiplying the 
semi-ordinate by the corresponding multiplier to find the area. 
These multipliers are placed at column 2 at the left, opposite 
the numbers of the ordinates. We have, therefore, under the 
heading of each water-line what we have termed the "functions 
of ordinates" and if these functions are added up, we shall 
obtain what we have termed the "function of area" 

Taking No. 3 W.L. as an instance, the "function" of its 
area is 144*10, and to convert this "function" into the actual 
area, we must multiply by one-third the common interval to 
complete Simpson's first rule, i.e. by \ X 7*i ; and also by 2 
to obtain the area of the water-plane on both sides of the ship. 
We should thus obtain the area of No. 3 W.L. 

144-10 x \ X 7-1 X 2 = 689-07 square feet 

The functions of the area of each water-plane are placed at 
the bottom of the columns, the figures being, starting with 
the L.W.L., 163-70, 155-36, 144-10, 128-74, 105-67, 87-27, 
and 60-97. To get the actual areas of each of the water-planes, 

1 In practice, it is advisable to put down the lengths of the semi- 
ordinates in some distinctive colour, such as red. 



72 Theoretical Naval Architecture. 

we should, as above, multiply each of these functions by \ X 
7* i X 2. Having the areas, we could proceed as on p. 26 to 
find the volume of displacement between No. i and No. 6 
water-lines, but we do not proceed quite in this way; we 
put the "functions of areas' 1 through Simpson's rule, and 
multiply afterwards by \ X T i X 2, the same result being 
obtained with much less work. Below the " functions of 
areas " are placed the Simpson's multipliers, and the products 
16370, 621*44, etc., are obtained. These products added up 
give 1 95 1 "83. This number is a function of the volume of 
displacement, this volume being given by first multiplying it 
by one-third the vertical interval, i.e. J x i ; and then by 
J X 7*1 X 2, as seen above. The volume of displacement, 
between No. i W.L. and No. 6 W.L. is therefore 

1951-83 x (J X i) X (J X 7'i) X 2 = 3079-5 cubic feet 

and the displacement in ) 3Q79'5 = 8rg8 tons 
tons (salt water) x ) 35 

We have thus found the displacement by dividing the 
volume under water by a series of equidistant horizontal planes ; 
but we could also find the displacement by dividing the under- 
water volume by a series of equidistant vertical planes, as we 
saw in Chapter I. This is done on the displacement sheet, 
an excellent check being thus obtained on the accuracy of the 
work. Take No. 4 section, for instance : its semi-ordinates, 
commencing with the L.W.L., are 6-40, 6-24, 5-90, 5-32, 4-30 
3-40, and 2-25 feet These ordinates are already put down 
opposite No. 4 ordinate. If these are multiplied successively 
by the multipliers, i, 4, 2, 4, i, 2, , and the sum of the 
functions of ordinates taken, we shall obtain the "function of 
area " of No. 4 section between the L.W.L. and 6 W.L. This 
is done in the table by placing the functions of ordinates 
immediately below the corresponding ordinate, the multiplier 
being given at the head of each column. We thus obtain a 
series of horizontal rows, and these rows are added up, the 
results being placed in the column headed u Function of areas? 
Each of these functions multiplied by one-third the common 
1 Thirty- five cubic feet of salt water taken to weigh one ton. 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 73 

interval, i.e. \ X i, and then by 2 for both sides, will give 
the areas of the transverse sections between the L.W.L. and 
6 W.L. ; but, as before, this multiplication is left till the end 
of the calculation. These functions of areas are put through 
Simpson's multipliers, the products being placed in the column 
headed " Multiples of areas''' This column is added up, giving 
the result 1951-83. To obtain the volume of displacement, 
we multiply this by (J X i) X 2 X ( X 7'i). It will be noticed 
that we obtain the number 1951-83 by using the horizontal 
water-lines and the vertical sections ; and this must evidently 
be the case, because the displacement by either method must 
be the same. The correspondence of these additions forms 
the check, spoken of above, of the accuracy of the work. We 
thus have the result that the volume of displacement from 
L.W.L. to 6 W.L. is 3079-5 cubic feet, and the displacement 
in tons of this portion 87-98 tons in salt water. This is termed 
the " Main solid," and forms by far the greater portion of the 
displacement. 

We now have to consider the portion we have left out 
below No. 6 water-plane. Such a volume as this is termed an 
" appendage? The sections of this appendage are given in the 
body-plan at the several stations. The form of these sections 
are traced off, and by the ordinary rules their areas are found 
in square feet. We have, therefore, this volume divided by a 
series of equidistant planes the same as the main solid, and we 
can put the areas of the sections through Simpson's rule and 
obtain the volume. This calculation is done on the left-hand 
side of the sheet, the areas being placed in column 3, and the 
functions of the areas in column 4. The addition of these 
functions is 49*99, and this multiplied by \ x 7'i gives the 
volume of the appendage in cubic feet, viz. 118-3; an d this 
volume divided by 35 gives the number of tons the appendage 
displaces in salt water, viz. 3-38 tons. The total displacement 
is thus obtained by adding together the, main solid and the 
appendage, giving 91-36 tons in salt water. The displacement 
in fresh water (36 cubic feet to the ton) would be 88-8 tons. 

The sheer drawing for this vessel as given on Plate I. was 
drawn to the frame line, i.e. to the moulded dimensions of the 



74 Theoretical Naval Architecture. 

ship ; but the actual ship is fuller than this, because of the outer 
bottom plating, and this plating will contribute a small amount 
to the displacement, but this is often neglected. Some sheer 
drawings, on the other hand, are drawn so that the lines include 
a mean thickness of plating outside the frame line, and when 
this is the case, the displacement sheet gives the actual dis- 
placement, including the effect of the plating. For a sheathed 
ship this is also true; in this latter case, the displacement 
given by the sheathing would be too great to be neglected. 
When the sheer drawing is drawn to the outside of sheathing, 
or to a mean thickness of plating, it is evident that the ship 
must be laid off on the mould loft floor, so that, when built, 
she shall have the form given by the sheer drawing. 

We now have to find the position of the centre of buoyancy 
both in a fore-and-aft and in a vertical direction. (It must be 
in the middle-line plane of the ship, since both sides are sym- 
metrical.) Take first the fore-and-aft position. This is found 
with reference to No. 6 station. The functions of the areas of 
the sections are 0-5, 23-055, etc., and in the column headed 
" Multiples of areas " we have these functions put through 
Simpson's multipliers. We now multiply these multiples by 
the number of intervals they respectively are from No. 6 station, 
viz. 5, 4, etc., and thus obtain a column headed " Moments." 
This column is added up for the fore body, giving 1505-43, and 
for the after body, giving 1913-02, the difference being 407-59 
in favour of the after body. To get the actual moment of the 
volume abaft No. 6 station, we should multiply this difference 
by (^ X i) for the vertical direction, (J x 7-1) for the fore-and- 
aft direction, and by 2 for both sides, and then by 7*1, since we 
have only multiplied by the number of intervals away, and not 
by the actual distances, or 47'59 X (J x i) X (J X 7*1) x 2 
X 7' i. The volume, as we have seen above, is given by 

1951*83 x (\ x i) x (i x 7-1) x 2 

The distance of the centre of gravity of the main solid from 
No. 6 station will be 

Moment - 1 - volume 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 75 

But on putting this down we shall see that we can cancel out, 
leaving us with 

W-59.XTI 8feet 

1951-83 

which is the distance of the centre of gravity of the main solid 
abaft No. 6 station. The distance of the centre of gravity of 
the appendage abaft No. 6 station is 4'o feet ; the working is 
shown on the left-hand side of the table, and requires no further 
explanation. 1 These results for the main solid and for the ap- 
pendages are combined together at the bottom ; the displacement 
of each in tons is multiplied by the distance of its centre of gravity 
abaft No. 6 station, giving the moments. The total moment 
is 143*73, and the total displacement is 91*36 tons, and this 
gives the centre of gravity of the total displacement, or what we 
term the centre of 'buoyancy -, C.B., 1*57 feet abaft No. 6 station. 
Now we have to consider the vertical position of the 
C.B., and this is determined with reference to the load water- 
line. For the main solid the process is precisely similar 
to that adopted for finding the horizontal position, with the 
exception that we take our moments all below the load water- 
plane, the number of intervals being small compared with the 
horizontal intervals. We obtain, as indicated on the sheet, the 
centre of gravity of the main solid at a distance of 2*21 feet 
below the L.W.L. For the appendage, we proceed as shown 
on the left-hand side of the sheet. When finding the areas of 
the sections of the appendage, we spot off as nearly as possible 
the centre of gravity of each section, and measure its distance be- 
low No. 6 W.L. If the sections happen to be triangles, this will, 
of course, be one-third the depth. These distances are placed in 
a column as shown, and the " functions of areas " are respec- 
tively multiplied by them, e.g. for No. 4 station the function of 
the area is 5*92, and this is multiplied by 0*22, the distance 
of the centre of gravity of the section of the appendage below 
No. 6 W.L. We thus obtain a column which, added up, gives 
a total of 13*78. To get the actual moment, we only have to 

1 For the vertical C.G. of the appendage Morrish's rule gives a good 
approximation. 



76 Theoretical Naval Architecture. 

multiply this by \ x yi. The volume of the appendage is 
49-99 X (5 X 7'i). So that the distance of the centre of 
gravity of the whole appendage below No. 6 W.L. is given 

by moment -T- volume, or l^J = 0-27 feet, and therefore the 

centre of gravity of the appendage is 5-27 feet below the 
L.W.L. The results for the main solid and for the appendage 
are combined together in the table at the bottom, giving the 
final position of the C.B. of the whole displacement as 2*32 
feet below the L.W.L. 

It will be of interest at this stage to test the two approxi- 
mations that were given on p. 65 for the distance of the C.B. 
below the L.W.L. The first was that this distance would be 
from ~ to Q of the mean draught to top of keel (i.e. the mean 
moulded draught). For this vessel the distance is 2^32 feet, 
and the mean moulded draught is 5' 9^", or 5*8 feet, and so 

we have the ratio -, or exactly ~. The second approxi- 

5' 8 
mation (Morrish's), p. 65, was 



(M) 



All these are readily obtainable from the displacement sheet, 
and if worked out its value is found to be 2*29 feet. This 
agrees fairly well with the actual result, 2*32 feet, the error 
being 3 in 232, or less than ij per cent. 

For large vessels a precisely similar displacement-sheet is 
prepared, but it is usual to add in the effect of other appen- 
dages besides that below the lowest W.L. Specimen calcula- 
tions are given on Tables II. and Ilia, at the end of the book. 

In the former the ordinates are to a mean thickness of 
plating. In the latter the moulded surface is used, and the 
displacement of the shell plating added as an appendage, being 
obtained by Denny's formula given on page 86. 

Graphic or Geometrical Method of calculating 
Displacement and Position of Centre of Buoyancy. 
There is one property of the curve, known as the "parabola of 
the second order" (see p. 6), that can be used in calculating 




Moments, Centre of Gravity, Centre of Buoyancy, etc. 77 

by a graphic method the area of a figure bounded by such a 
curve. Let BFC, Fig. 37, be a curve bounding the figure 
ABCD, and suppose the curve is a "parabola of the second 
order" Draw the ordinate EF 
midway between AB and DC ; then 
the following is a property of the 
curve BFC : the area of the seg- 
ment BCF is given by two thirds 
the product of the deflection GF 
and the base AD, or 

Area BCF = f X GF x AD 
Make GH = |GF. Then- 
Area BCF = GH X AD 

Now, the area of the trapezoid ABCD is given by 
AD x EG, and consequently 

The area ADCFB = AD x EH 

Thus, if we have a long curvilinear, we can' divide it up 
as for Simpson's first rule, and set off on each of the inter- 
mediate ordinates two-thirds the deflection of the curve above 
or below the straight line joining the extremities of the dividing 
ordinates. Then add together on a strip of paper all such 
distances as EH right along, and the sum multiplied by the 

1 This property may be used to prove the rule known as Simpson's first 
rule. Call AB, EF, DC respectively y y z , y 3 . Then we have 



E. 
FIG. 37. 



EG 



andFG=j 2 - EG 



EH = EG + GH 



^'tZJ'LZ. 

3 



and calling AE = h, we have 

Area ADCFB = |fo + 4J 2 +J' 3 ) 
which is the same expression as given by Simpson's first rule. 



78 Theoretical Naval Architecture. 

distance apart of the dividing ordinates, as AD, will give the 
area required. Thus in Fig. 38, AB is divided into equal parts 




FIG 



as shown. D and E are joined, also E and C ; MO is set 
off = |HM, and NP is set off = f NK. Then- 

Area ADEF = AF X GO 
and area FECB = FB X LP 
and the whole area ABCD = AF X (GO + LP) 

We can represent the area ABCD by a length equal to 
GO + LP on a convenient scale, if we remember that this length 
has to be multiplied by AF to get the area. This principle can 
be extended to finding the areas of longer figures, such as 
water-planes, and we now proceed to show how the displacement 
and centre of buoyancy of a ship can be determined by its use. 
The assumption we made at starting is supposed to hold good 
with all the curves we have to deal, i.e. that the portions 
between the ordinates are supposed to be "parabolas of the 
second order" This is also the assumption we make when 
using Simpson's first rule for finding displacement in the ordi- 
nary way. 

Plate I. represents the ordinary sheer drawing of a vessel, 
and the underwater portion is divided by the level water-planes 
shown by the half-breadth plan. The areas of each of these 
planes can be determined graphically as above described, the 
area being represented by a certain length obtained by the 
addition of all such lengths as GO, etc., Fig. 38, the interval 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 79 

being constant for all the water-planes. Let AB, Fig. 39, be 
set vertically to represent the extreme moulded draught of the 
vessel. Draw BC at right angles to AB, to represent on a 
convenient scale the area of the L.W.L. obtained as above. 
Similarly, DE, FG are set out to represent on the same scale 
the areas of water-planes 2 and 3, and so on for each water- 
plane. A curve drawn through all such points as C, E, and G 



L.W.I 



2.W.L. 




FIG. 39. 

will give a " curve of areas of wafer-planes" Now, the area of this 
curve up to the L.W.L. gives us the volume of displacement up 
to the L.W.L., as we have seen in Chapter I., and we can readily 
find the area of the figure ABCEG by the graphic method, and 
this area will give us the displacement up to the L.W.L. 
Similarly, the area of ADEG will give the displacement up to 
2 W.L., and so on. Therefore set off BL to represent on a 



8o Theoretical Naval Architecture. 

convenient scale the area of the figure ABCE, DK on the 
same scale to represent the area ADEG, and so on. Then 
a curve drawn through all such points as L, K will give us a 
" curve of displacement" and the ordinate of this curve at any 
draught will give the displacement at that draught, BL being 
the load displacement. 

We now have to determine the distance of the centre 
of buoyancy below the L.W.L., and to find this we must get 
the moment of the displacement about the L.W.L. and 
divide this by the volume of displacement below the L.W.L. 
We now construct a curve, BPMA, such that the ordinate at 
any draught represents the area of the water-plane at that 
draught multiplied by the depth of the water-plane below the 
L.W.L. Thus DP represents on a convenient scale the area 
of No. 2 water-plane multiplied by DB, the distance below the 
L.W.L. The ordinate of this curve at the L.W.L. must evi- 
dently be zero. This curve is a curve of " moments of areas 
of water-planes" about the L.W.L. The area of this curve up 
to the L.W.L. will evidently be the moment of the load dis- 
placement about the L.W.L., and thus the length BR is set out 
to equal on a convenient scale the area of BPMA. Similarly, 
DS is set out to represent, on the same scale, the area of 
DPMA, and thus the moment of the displacement up to 2 W.L. 
about the L.W.L. These areas are found graphically as in the 
preceding cases. Thus a curve RSTA can be drawn in, and 
BR -' BL, or moment of load displacement about L.W.L. -7- 
load displacement, gives us the depth of the centre of buoyancy 
for the load displacement below the L.W.L. 

Exactly the same course is pursued for finding the displace- 
ment and the longitudinal position of the centre of buoyancy, 
only in this case we use a curve of areas of transverse sections 
instead of a curve of areas of water-planes, and we get the 
moments of the transverse areas about the middle ordinate. 
Fig. 40 gives the forms the various curves take for the fore 
body. AA is the " curve of areas of transverse sections ; " BB 
is the " curve of displacement " for the fore body, OB being the 
displacement of the fore body. CC is the curve of " moments 
of areas of transverse sections " about No. 6 ordinate ; DD is 




Moments, Centre of Gravity, Centre of Buoyancy, etc. 8 1 

the curve of "moment of displacement" about No. 6 ordinate, 
OD being the moment of the fore-body displacement about 
No. 6 ordinate. Similar curves can be drawn for the after 
body, and the dif- 
ference of the mo- 
ments of the fore and 
after bodies divided 
by the load displace- 
ment will give the 
distance of the centre 
of buoyancy forward 
or aft of No. 6 ordi- 
nate, as the case may be. The total displacement must be the 
same as found by the preceding method. 1 

Method of finding Areas by Means of the Plani- 
meter. This instrument is frequently employed to find the 
area of plane curvilinear figures, and thus the volume of dis- 
placement of a vessel can be determined. One form of the 
instrument is shown in diagram by Fig. 41. It is supported at 
three places : first, by a weighted pin, which is fixed in position 
by being pressed into the paper ; second, by a wheel, which 
actuates a circular horizontal disc, the wheel and disc both 
being graduated ; and third, by a blunt pointer. The instru- 
ment is placed on the drawing, the pin is fixed in a convenient 
position, and the pointer is placed on a marked spot A on the 
boundary of the curve of which the area is required. The 
reading given by the wheel and disc is noted. On passing 
round the boundary of the area with the pointer (the same way 
as the hands of a clock) back to the starting-point, another 
reading is obtained. The difference of the two readings is 
proportional to the area of the figure, the multiplier required to 
convert the difference into the area depending on the instru- 
ment and on the scale to which the figure is drawn. Particu- 
lars concerning the necessary multipliers are given with the 
instrument ; but it is a good practice to pass round figures of 
known area to get accustomed to its use. 

1 This method may be considerably simplified by using the property 
of the curve of displacement given on p. 66. 

G 



82 Theoretical Naval Architecture. 

By the use of the planimeter the volume of displacement of 
a vessel can very readily be determined. The body plan is 
taken, and the L.W.L. is marked on. The pointer of the in- 
strument is then passed round each section in turn, up to the 
L.W.L., the readings being tabulated. If the differences of the 
readings were each multiplied by the proper multiplier, we 
should obtain the area of each of the transverse sections, and 
so, by direct application of Simpson's rules, we should find the 




POINTER. 



FIG. 



required volume of displacement. Or we could put the actual 
difference of readings through Simpson's multipliers, and 
multiply at the end by the constant multiplier. 

It is frequently the practice to shorten the process as 
follows : The body-plan is arranged so that Simpson's first rule 
will be used, i.e. an odd number of sections is employed. 
The pointer is passed round the first and last sections, and 
the reading is recorded. It is then passed round all the even 
sections, 2, 4, 6, etc., and the reading is recorded. Finally, 
it is passed round all the odd sections except the first and last, 
viz. 3, 5> 7> et c., and the reading is put down. The differences 
of the readings are found and put down in a column. The 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 83 



first difference is multiplied by i, the next difference is multi- 
plied by 4, and the last by 2. The sum of these products is 
then multiplied for Simpson's first rule, and then by the proper 
multiplier for the instrument and scale used. The work can 
conveniently be arranged thus : 



Numbers of 
sections. 


Readings. 


Differences 
of readings. 


Simpson's 
multipliers. 


Products. 


Initial reading 


5 I2 4 





__ 


__ 


I, 21 


S.S^O 


2 3 6 


I 


236 


2, 4,6, 8, 10, ) 










12, 14, 16, [ 


18,681 


13,321 


4 


53,284 


18, 20 ... ) 










3, 5> 7, 9, ii, 1 










13, 15, !7 [ 


31,758 


13,077 


2 


26,154 


19 ) 











79,674 

The multiplier for the instrument and scale of the drawing 
used and to complete the use of Simpson's first rule is ; so 
that the volume of displacement is 79,674 X cubic feet, and 
the displacement in tons is 79,674 x X A = 2732 tons. 

There are two things to be noticed in the use of the plani- 
meter: first, it is not necessary to set the instrument to the 
exact zero, which is somewhat troublesome to do ; and second, 
the horizontal disc must be watched to see how many times 
it makes the complete revolution, the complete revolution 
meaning a reading of 10,000. 

It is also possible to find the vertical position of the centre 
of buoyancy by means of the planimeter. By the method 
above described we can determine the displacement up to each 
water-line in succession, and so draw in on a convenient scale 
the ordinary curve of displacement. Now we can run round 
this curve with the planimeter and find its area. This area 
divided by the top ordinate (i.e. the load displacement) will 
give the distance of the centre of buoyancy below the load-line 
(see p. 67). 

To find the centre of buoyancy in a fore-and-aft direction, 
it is necessary to tabulate the differences for each section, 
and treat these differences in precisely the same way as the 



Theoretical Naval Architecture. 



" functions of areas of vertical sections " are treated in the 
ordinary displacement sheet. 

Method of approximating to the Area of the 
Wetted Surface 1 by "Kirk's" Analysis. The ship is 
assumed to be represented by a block model, shaped as 
shown in Fig. 42, formed of a parallel middle body and a 




r 




FIG. 48. 

tapered entrance and run which are taken as of equal length. 
The depth of the model is equal to the mean draught, and the 
length of the model is equal to the length of the vessel. The 
breadth is not equal to the breadth of the vessel, but is equal 
to area of immersed midship section -f- mean draught. The 
displacement of the model is made equal to that of the vessel. 
We then have 
Volume of displace-] _ y s 
ment ) 

= AG X area of midship section 
V 

~ area of midship section 
.'. length of entrance} . V 

> = length Of Ship - c r^r-. 

or run area of midship section 

V 
B' X D 

1 The area of wetted surface can be closely approximated to by putting 
a. curve of girths (modified for the slope of the level lines, see p. 228) 
through Simpson's rule. 



Moments. Centre of Gravity, Centre of Buoyancy, etc. 85 

where L = length of ship ; 

B' = breadth of model ; 

D = mean draught. 

Having found these particulars, the surface of the model 
can be readily calculated. 

Area of bottom = AG X B' 
Area of both sides = 2(GH -f- 2AE) x mean draught 

The surface of a model formed in this way approximates 
very closely to the actual wetted surface of the vessel. It is 
stated that in very fine ships the surface of the model exceeds 
the actual wetted surface by about 8 per cent., for ordinary 
steamers by about 3 per cent., and for full ships by 2 per cent. 

By considering the above method, we may obtain an 
approximate formula for the wetted surface 

V 
Area of bottom = ~ 

Area of sides = 2L/D 
where L' is the length along ADCB. Then 

V 
Surface = 2L'D -f- ^ 

This gives rather too great a result, as seen above ; and if 
we take 

V 
Surface = 2LD + ^ 

we shall get the area of the wetted surface slightly in excess, 
but this will allow for appendages, such as keels, etc. 

Since V = k . LED, where k is the block coefficient of 
displacement, we may write 

. Surface = 2LD + k . LB 

Approximate Formulae for finding Wetted Surface. 
Mr. Denny gives the following formula for the area of 
wetted surface : 



86 



Theoretical Naval Architecture. 



i- 7 LD + ^ 



which is seen to be very nearly that obtained above. 

Mr. Taylor, in his work on " Resistance and Propulsion 
of Ships," gives the following formula : 

15-5 v/WL 

where W is the displacement in tons. 

The following formula for the wetted surface is used at the 
experimental tank at Haslar 



One of these formulae can be used to find the area of 
wetted surface to calculate the displacement of the skin plating, 
as is necessary when the sheer drawing is drawn to the 
moulded surface of the ship. See Brown's Displacement Sheet 
in Appendix, in which Denny's approximation is used. 



EXAMPLES TO CHAPTER IL 



I. A ship has the following weights placed on board : 



20 tons 

45 

15 

60 

40 

30 



100 feet before amidships 

80 

40 ,, 

50 feet abaft ,, 

80 
no 



Show that these weights will have the same effect on t. r m.u of the ship 
as a single weight of 210 tons placed 15$ feet abaft amidships. 

2. Six weights are placed on a drawing-board. The weights are 3, 4, 
5, 6, 7, 8 Ibs. respectively. Their distances from one edge are 5, 4|, 4, 3}, 
3, 2 feet respectively, and from the edge at right angles, &, \, I, i$, 2, 2$, 
feet respectively. The drawing-board weighs 6 Ibs., and is 6 feet long 
and 3 feet broad. Find the position where a single support would need 
to be placed in order that the board should remain horizontal. 

Ans. 3*27 feet from short edge, 1-58 feet from long edge. 



Moments, Centre of Gravity \ Centre of Buoyancy, etc. 87 

3. An area bounded by a curve and a straight line is divided by ordinates 
4 feet apart of the following lengths : o, 12*5, 14-3, 15*1, 15-5, 15-4, 14*8, 
14*0, o feet respectively. Find 

(1) Area in square feet. 

(2) Position of centre of gravity relative to the first ordinate. 

(3) Position of the centre of gravity relative to the base. 

Ans. (i) 423 square feet ; (2) 16*27 feet ; (3) 7*24 feet. 

4. A triangle ABC has its base EC 15 feet long, and its height 25 
feet. A line is drawn 10 feet from A parallel to the base, meeting AB and 
AC in D and E. Find the distance of the centre of gravity of DBCE 
from the apex. 

Ans. i8'57 feet. 

5. The semi -ordinates of a water-plane in feet, commencing from the 
after end, are 5'2, ICT2, 14-4, 17-9, 2O'6, 227, 243. 25-5, 26*2, 26*5, 
26-6, 26-3, 25-4, 23-9, 2i'8, i8'8, 15-4, 11-5, 7-2, 3-3, 2-2. The distance 
apart is 15 feet. Find the area of the water -plane, and the position of the 
centre of gravity in relation to the middle ordinate. 

Ans. 11,176 square feet; 10-15 feet abaft middle. 

6. Find the area and transverse position of the centre of gravity of 
"half" a water-line plane, the ordinates in feet being 0*5, 6, 12, 16, 12, 10, 
and 0*5 respectively, the common interval being 15 feet. 

Ans. 885 square feet ; 6*05 feet. 

7. The areas of sections 17' 6" apart through a bunker, commencing 
from forward, are 65, 98, 123, 137, 135, 122, 96 square feet respectively. 
The length of bunker is 100 feet, and its fore end is i' 6" forward of the 
section whose area is 65 square feet. Draw in a curve of sectional areas, 
and obtain, by using convenient ordinates, the number of cubic feet in the 
bunker, and the number of tons of coal it will contain, assuming that 43 cubic 
feet of coal weigh i ton. Find also the position of the C.G. of the coal 
relative to the after end of the bunker. 

Ans. 272 tons ; 46 J feet from the after end. 

8. The tons per inch in salt water of a vessel at water-lines 3 feet 
apart, commencing with the L.W.L., are 31*2, 3o - o, 28-35, 2 6'2i, 23*38, 
19-5, I2'9. Find the displacement in salt and fresh water and the position 
of the C.B. below the L.W.L., neglecting the portion below the lowest 
W.L. Draw in the tons per inch curve for salt water to a convenient scale, 
and estimate from it the weight necessary to be taken out in order to lighten 
the vessel 2' 3^" from the L.W.L. The mean draught is 20' 6". 

Ans. 5405 tons ; 5255 tons ; 8'OI feet ; 847 tons. 

9. In the preceding question, calling the L.W.L. i, find the displacement 
up to 2 W.L., 3 W.L., and 4 W.L., and draw in a curve of displacement 
from the results you obtain, and check your answer to the latter part of the 
question. 

10. The tons per inch of a ship's displacement at water-lines 4 feet 
apart, commencing at the L.W.L., are 44*3, 427, 40' 5, 37- 5, 33'3- Find 
number of tons displacement, and the depth of C.B. below the top W.L. 

Ans. 7670 tons ; 7 '6 feet. 

11. The ship in the previous question has two water-tight transverse 
bulkheads 38 feet apart amidship, and water-tight flats at 4 feet below and 
3 feet above the normal L.W.L. If a hole is made in the side 2 feet 
below the L.W.L., how much would the vessel sink, taking the breadth 
of the L.W.L. amidships as 70 feet? Indicate the steps where, owing to 
insufficient information, you are unable to obtain a perfectly accurate result. 

Ans. 8 inches. 

12. The areas of transverse sections of a coal -bunker 19 feet apart are 



88 Theoretical Naval Architecture. 

respectively 63-2, 93*6, 121 '6, io8'8, 94-8 square feet, and the centres 
of gravity of these sections are ID'S, ir6, I2'2, 117, 11*2 feet respectively 
below the L.W.L. Find the number of tons of coal the bunker will hold, 
and the vertical position of its centre of gravity (44 cubic feet of coal to the 
ton). 

Ans. 174-3 tons; ii'68 feet below L.W.L. 

13. A vessel is 180 feet long, and the transverse sections from the load 
water-line to the keel are semicircles. Find the longitudinal position of 
the centre of buoyancy, tke ordinates of the load water-plane being I, 5, 13, 
15, 14, 12, and 10 feet respectively. 

Ans, 1 06 '2 feet from the finer end. 

14. Estimate the distance of the centre of buoyancy of a vessel below 
the L.W.L., the vessel having 22' 6" mean moulded draught, block co- 
efficient of displacement 0x55, coefficient of fineness of L.W.L. O'7 (use 
Morrish's formula, p. 65). 

Ans. 9-65 feet. 

15. A vessel of 2210 tons displacement, 13' 6" draught, and area 
of load water-plane 8160 square feet, has the C.B., calculated on the dis- 
placement sheet, at a distance of 5*43 feet below the L.W.L. Check this 
result. 

1 6. The main portion of the displacement of a vessel has been calculated 
and found to be 10,466 tons, and its centre of gravity is 10-48 feet below 
the L.W.L., and 5-85 feet abaft the middle ordinate. In addition to this, 
there are the following appendages : 

tons. 

Below lowest W.L. 263, 24*8 ft. below L.W.L., 4-4 ft. abaft mid. ord. 

Forward 5, I2'o 202 ft. forward of mid 

ord. 

Stern ... ... 16, 2'8 ,, 201 ft. abaft mid. ord. 

Rudder 16, 17-5 200 ,, 

Bilge keels ... 20, 20 o ,, 

Shafting, etc. ... 18, 15 ,, ,, 140 ,, ,, 

Find the total displacement and position of the centre of buoyancy. 

Ans. 10,804 tons ; C.B. 6'5 abaft mid. ord., IO'86 ft. below L.W.L. 

17. The displacements of a vessel up to water-planes 4 feet apart 
are 10,804, 8612, 6511, 4550, 2810, 1331, and 263 tons respectively. The 
draught is 26 feet. Find the distance of the centre of buoyancy below the 
load water-line. 

Ans. ro'9 feet nearly. 

18. The load displacement of a ship is 5000 tons, and the centre of 
buoyancy is 10 feet below the load water-line. In the light condition the 
displacement of the ship is 2000 tons, and the centre of gravity of the layer 
between the load and light lines is 6 feet below the load-line. Find the 
vertical position of the centre of buoyancy below the light line in the light 
condition. 

Ans. 4 feet, assuming that the C.G. of the layer is at half its depth. 

19. Ascertain the displacement and position of the centre of buoyancy 
of a floating body of length 140 feet, depth 10 feet, the forward section 
being a triangle 10 feet wide at the deck and with its apex at the keel, and 
the after section a trapezoid 20 feet wide at the deck and 10 feet wide at 
the keel, the sides of the vessel being plane surfaces; draught of water 
may be taken as 7 feet. 

Ans. 238 tons ; 56*3 feet before after end, 3 feet below water-line. 

20. Show by experiment or otherwise that the centre of gravity of 8 



Moments, Centre of Gravity, Centre of Buoyancy, etc. 89 

quadrant of a circle 3 inches radius is I '8 inches from the right angle of 
the quadrant. 

21. A floating body has a constant triangular section, vertex down- 
wards, and has a constant draught of 12 feet in fresh water, the breadth at 
the water-line being 24 feet. The keel just touches a quantity of mud of 
specific gravity 2. The water-level now falls 6 feet. How far will the 
body sink into the mud ? 

Ans. 4 feet 11^ inches. * 

22. Show that the C.G. of a trapezoid, as ABCD, Fig. 5, is distant 
I\~L.) from the middle of the tength h. The C.G. must be on a line 

joining the centres of the parallel sides. Thus the position of the C.G. 
is fully determined. This will also apply to a figure in which the parallel 
sides are not perpendicular to one of the other sides. 

23. Apply Morrish's rule to find the C.G. of a semicircle, and state the 
error involved if the semicircle is 20 feet radius. 

Ans. 0*19 foot. 

24. For the lower appendage of the ship displacement sheet, Table I., 
find the vertical C.G. by using Morrish's rule. 

25. Describe the process of finding the area and position of the C.G. 
of a plane figure by radial integration, and apply it to find these in the case 
of a rectangle 8 feet wide and 1 2 feet long. 

26. A vessel is 300 feet x 36$ teet X 13^ feet draught, 2135 tons dis- 
placement. Find the area of wetted surface by each of the formula 
given on p. 86. 

Ans., Denny, 12,421 square feet. 
Taylor, 12,400 square feet. 
Froude, 12,350 square feet. 

The student is advised to take a sheer drawing and obtain a close 
approximation to the wetted surface by putting the half girths to water- 
line through Simpson's rule. Then to apply the above formulae and see 
what the comparative results are. 

27. Show, by means of the result in question 22, that the C G. of a 
trapezoid in relation to the parallel sides is given by the construction of 
Fig. 126. 

28. Having given the mean ordinate/of a trapezoid and the distance 
x of its C.G. from the larger end, show that the lengths of the parallel 
sides are 

/ 6Ar\ , 

(4/--f)and 

where h is the length. 

1 This example is worked out at the end of Appendix A. 



CHAPTER TIL 



CONDITIONS OF EQUILIBRIUM, TRANSVERSE MET A* 
CENTRE, MOMENT OF INERTIA, TRANSVERSE BM, 
INCLINING EXPERIMENT, METACENTRIC HEIGHT, 
ETC. 

Trigonometry. The student of this subject will find it a 
( distinct advantage, especially when dealing with the question 
of stability, if he has a knowledge of some of the elementary 
portions of trigonometry. The following are some properties 
which should be thoroughly grasped : 

Circular Measure of Angles. The degree is the unit gene- 
rally employed for the measurement of angles. A right angle 

is divided into 90 equal 
parts, and each of these 
parts is termed a "de- 
gree." If two lines, as 
OA, OB, Fig. 43, are 
inclined to each other, 
forming the angle AOB, 
and we draw at any radius 
OA an arc AB from the 
centre O, cutting OA, 
OB in A and B, then 




O. A. 

Ftc. 43- 

length of arc AB -f- radius OA is termed the circular measure 
of the angle AOB. Or, putting it more shortly 



arc 



Circular measure = - 



The circular measure of four right ) 
angles, or 360 degrees 



radius 

circumference of a circle 



radius 



27T 



Conditions of Equilibrium, Transverse Metacentre, etc. 



The circular measure of a right angle / = 

Since 360 degrees = 2?r in circular measure, then the angle 
whose circular measure is unity is 

360 

= 57'3 degrees 

The circular measure of i degree is 7- = 0*01745, and 

thus the circular measure of any angle is found by multiplying 
the number of degrees in it by 

Trigonometrical Ratios? 
etc. Let BOC, Fig. 44, be 
any angle ; take any point P 
in one of the sides OC, and 
draw PM perpendicular to 
OB. Call the angle BOC, 6>. 2 

PM is termed the perpen- 
dicular. 

OM is termed the base. 




OP is termed the hypo- 
tenuse. O. 

Then 




I-A 



BASE. 

FIG. 44. 



PM perpendicular 

7^ = - 1 - = sine 0, usually written sin 6 

OP hypotenuse 

OM base 

7 ~- = . = cosine 6, usually written cos 6 

OP hypotenuse 

PM perpendicular . ,. 
= C r - = tangent 6, usually written tan 6 

These ratios will have the same value wherever P is taken 
on the line OC. 

1 An aid to memory which is found of assistance by many in learning 
these ratios is 

Sin /<rrplexes ^y/ocrites t 

Cos of base /y/ocrisy. 
9 6 is a Greek letter (theta) often used to denote an angle. 



9 2 



Theoretical Naval Architecture 



We can write sin = 



cos = 



and also tan 6 = 



hyp- 
base 

hyp. 

sin 

cos e 



There are names for the inversions of the above ratios, 
which it is not proposed to use in this work. 

For small angles, the value of the angle in circular 
measure is very nearly the same as the values of sin and 
tan &. This will be seen by comparing the values of 0, sin 0, 
and tan for the following angles : 



Angle in 
degrees. 


Angle in 
circular 
measure. 


Sin A 


,Tan 6. 


2 


0'0349 


0-0349 


0-0349 


4 


0-0698 


0-0697 


0-0699 


6 


0-1047 


0-1045 


0-1051 


8 


0-1396 


0-1392 


0-I405 


1C 


0-1745 


0-1736 


0-1763 



Up to 10 they have the same values to two places of 
decimals, and for smaller angles the agreement in value is 
closer still. 

Tables of sines, cosines, and tangents of angles up to 90 
are given in Appendix B. 

Conditions that must hold in the Case of a Vessel 
floating freely, and at Rest in Still Water. We 
saw in Chapter I. that, for a vessel floating in still water, 
the weight of the ship with everything she has on board must 
equal the weight of the displaced water. To demonstrate this, 
we imagined the cavity left by the ship when lifted out of the 
water to be filled with water (see Fig. 17). Now, the upward 
support of the surrounding water must exactly balance the weight 
of the water poured in. This weight may be regarded as acting 
downwards through its centre of gravity, or, as we now term 
it, the centre of buoyancy. Consequently, the upward support 



Conditions of Equilibrium, Transverse Metacentre, etc. 93 

of the water, or the buoyancy, must act through the centre of 
buoyancy. All the horizontal pressures of the water on the 
surface of the ship must evidently balance among themselves-. 
We therefore have the following forces acting upon the ship : 

(1) The weight acting downwards through the C.G. ; 

(2) The upward support of the water, or, as it is termed, 

the buoyancy, acting upwards through the C.B. ; 
and for the ship to be at rest, these two forces must act in the 
same line and counteract each other. Consequently, we also 
have the following condition : 

The centre of gravity of the ship, with everything she has on 
board, must be in the same vertical line as the centre of buoyancy. 

If a rope is pulled at both ends by two men exerting the 
same strength, the rope will evidently remain stationary; and 
this is the case with a ship floating freely and at rest in still 
water. She will have no tendency to move of herself so long 
as the C.G. and the C.B. are in the same vertical line. 

Definition of Statical Stability. The statical 
stability of a vessel is the tendency she has to return to the 
upright when inclined away from that position. It is evident 
that under ordinary conditions of service a vessel cannot 
always remain upright ; she is continually being forced away 
from the upright by external forces, such as the action of 
the wind and the waves. It is very important that the ship 
shall have such qualities that these inclinations that are forced 
upon her shall not affect her safety ; and it is the object of the 
present chapter to discuss how these qualities can be secured 
and made the subject of calculation so far as small angles of 
inclination are concerned. 

A ship is said to be in stable equilibrium for a given direc- 
tion of inclination if, on being slightly inclined in that direction 
from her position of rest, she tends to return to that position. 

A ship is said to be in unstable equilibrium for a given 
direction of inclination if, on being slightly inclined in that 
direction from her position of rest, she tends to move away 
farther from that position. 

A ship is said to be in neutral or indifferent equilibrium 
for a given direction of inclination if, on being slightly inclined 



94 



Ttieoretical Naval Architecture. 



in that direction from her position of rest, she neither tends 
to return to nor move farther from that position. 

These three cases are represented by the case of a heavy 
sphere placed upon a horizontal table. 

1. If the sphere is weighted so that its C.G. is out of the 
centre, and the C.G. is vertically below the centre, it will be 
in stable equilibrium. 

2. If the same sphere is placed so that its C.G. is vertically 
above the centre, it will be in unstable equilibrium. 

3. If the sphere is formed of homogeneous material so that 
its C.G. is at the centre, it will be in neutral or indifferent 
equilibrium. 

Transverse Metacentre. We shall deal first with 
transverse inclinations, because they are the more important, 
and deal with inclinations in a longitudinal or fore-and-aft 
direction in the next chapter. 

Let Fig. 45 represent the section of a ship steadily inclined 




STABLE. 



Fir,. 45. 

at a small angle from the upright by some external force, 
such as the wind. The vessel has the same weight before and 
after the inclination, and consequently has the same volume 
of displacement. We must assume that no weights on board 
shift, and consequently the centre of gravity remains in the 
same position in the ship. But although the total volume of 



Conditions of Equilibrium, Transverse Metacentre, etc. 95 

displacement remains the same, the shape of this volume 
changes, and consequently the centre of buoyancy will shift 
from its original position. In the figure the ship is repre- 
sented by the section, WAL being the immersed section 
when upright, WL being the position of the water-line on 
the ship. On being inclined, WL' becomes the water-line, 
and WAL' represents tbe immersed volume of the ship, which, 
although different in shape, must have the same volume as the 
original immersed volume WAL. 

The wedge-shaped volume represented by WSW, which 
has come out of the water, is termed the " emerged" or " out" 
wedge. The wedge-shaped volume represented by LSL', 
which has gone into the water, is termed the "immersed" or 
" in " wedge. Since the ship retains the same volume of 
displacement, it follows that the volume of the emerged wedge 
WSW is equal to the -volume of the immersed wedge LSL'. 
It is only for small angles of inclination that the point S, 
where the water-lines intersect, falls on the middle line of the 
vessel. For larger angles it moves further out, as shown in 
Fig. 77. 

Now consider the vessel inclined at a small angle from 
the upright, as in Fig. 45. The new volume of displacement 
WAL' has its centre of buoyancy in a certain position, say B'. 
This position might be calculated from the drawings in the 
same manner as we found the point B, the original centre of 
buoyancy ; but we shall see shortly how to fix the position of 
the point B' much more easily. 

B' being the new centre of buoyancy, the upward force of 
the buoyancy must act through B', while the weight of the ship 
acts vertically down through G, the centre of gravity of the 
ship. Suppose the vertical through B' cuts the middle line of 
the ship in M ; then we shall have two equal forces acting on 
the ship, viz. 

(1) Weight acting vertically down through the centre of 

gravity. 

(2) Buoyancy acting vertically up through the new centre 

of buoyancy. 
But they do not act in the same vertical line. Such a system 



9 6 



Theoretical Naval Architecture. 



of forces is termed a couple. Draw GZ perpendicular to the 
vertical through B'. Then the equal forces act at a distance 
from each other of GZ. This distance is termed the arm of 
the couple, and the moment of the couple is W X.GZ. On 
looking at the figure, it is seen that the couple is tending to 
take the ship back to the upright If the relative positions 
of G and M were such that the couple acted as in Fig. 46, 
the couple would tend to take the ship farther away from the 
upright ; and again, if G and M coincided, we should have the 
forces acting in the same vertical line, and consequently no 




UNSTABLE. / 



FIG. 46. 

couple at all, and the ship would have no tendency to move 
either to the upright or away from it. 

We see, therefore, that for a ship to be in stable equilibrium 
for any direction of inclination, it is necessary that the point 
M be above the centre of gravity of the ship. This point M 
is termed the metacentre. We now group together the three 
conditions which must be fulfilled in order that a ship may 
float freely and at rest in stable equilibrium 

(i) The weight of water displaced must equal the total 
weight of the ship (see p. 23). 



Conditions of Equilibrium, Transverse Metacentre, etc. 97 

(2) The centre of gravity of the ship must be in the same 
vertical line as the centre of gravity of the displaced water 
(centre of buoyancy) (see p. 93). 

(3) The centre of gravity of the ship must be below the 
metacentre. 

For small transverse inclinations, M is termed the transverse 
metacentre^ which we may accordingly define as follows : 

For a given plane of flotation of a vessel in the upright 
condition, let B be the centre of buoyancy, and BM the vertical 
through it. Suppose the vessel inclined transversely through 
a very small angle, retaining the same volume of displacement, 
B' being the new centre of buoyancy, and B'M the vertical 
through it, meeting BM in M. Then this point of intersection, 
M, is termed the transverse metacentre. 

There are two things in this definition that should be noted : 
(i) the angle of inclination is supposed very small, and (2) the 
volume of displacement remains the same. 

It is found that, for all practical purposes, in ordinary ships 
the point M does not change in position for inclinations up to 
as large as 10 to 15; but beyond this it takes up different 
positions. 

We may now say, with reference to a ship's initial stability 
or stability in the upright condition 

(1) If G is below M, the ship is in stable equilibrium. 

(2) If G is above M, the ship is in unstable equilibrium. 

(3) If G coincides with M, the ship is in neutral or in- 
different equilibrium. 

We thus see how important the relative positions of the 
centre of gravity and the transverse metacentre are as affecting 
a ship's initial stability. The distance GM is termed the 
transverse metacentric height^ or, more generally, simply the 
metacentric height. 

We have seen that for small angles M remains practically 
in a constant position, and consequently we may say GZ 
= GM . sin 6 for angles up to 10 to 15, say. GZ is the 
arm of the couple, and so we can say that the moment of the 
couple is 

W x GM . sin B 

H 



9$ Theoretical Naval Architecture. 

If M is above G, this moment tends to right the ship, and 
we may therefore say that the moment of statical stability at the 
angle is 

W X GM . sin 

This is termed the metacentric method of determining a 
vessel's stability. It can only be used at small angles of 
inclination to the upright, viz. up to from 10 to 15 degrees. 

Example. A vessel of 14,000 tons displacement has a metacentric 
height of 3^ feet. Then, if she is steadily inclined at an angle of 10, the 
tendency she has to return to the upright, or, as we have termed it, the 
moment of statical stability, is 

14,000 x 3 '5 X sin 10 = 8506 foot-tons 

We shall discuss later how the distance between G and M, 
or the metacentric height, influences the behaviour of a ship, 
and what its value should be in various cases ; we must now 
investigate the methods which are employed by naval archi- 
tects to determine the distance for any given ship. 

There are two things to be found, viz. (i) the position of 
G, the centre of gravity of the vessel ; (2) the position of M, 
the transverse metacentre. 

Now, G depends solely upon the vertical distribution of the 
weights forming the structure and lading of the ship, and the 
methods employed to find its position we shall deal with 
separately ; but M depends solely upon the form of the ship, 
and its position can be determined when the geometrical form 
of the underwater portion of the ship is known. Before we 
proceed with the investigation of the rules necessary to do this, 
we must consider certain geometrical principles which have to. 
be employed. 

Centre of Flotation. If a floating body is slightly 
inclined so as to maintain the same volume of displacement, 
the new water-plane must pass through the centre of gravity of 
the original water-plane. In order that the same volume of 
displacement may be retained, the volume of the immersed 
wedge SLL 1} Fig 47, must equal the volume of the emerged 
wedge SWWj. Call y an ordinate on the immersed side, and 
y' an ordinate on the emerged side of the water-plane. Then 



Conditions of Equilibrium, Transverse Metacentre, etc. 99 

the areas of the sections of the immersed and emerged wedges 
are respectively (since LI^ = y . dO, WWi = / . dO t dQ being 
the small angle of inclination) . 

*/.<#, ' *(/)' <# 
and using the notation we have already employed 

Volume of immersed wedge = %fy z . dO .dx 

emerged =Wf.M.dx 
ind accordingly 

.dd.dx 



SECTION. - 



or 

But J/T* . dx is the moment of the immersed portion of the 

water-plane about the intersection, 

and i/(y) 2 . dx is the moment of 

the emerged portion of the water- 

plane about the intersection (see 

p. 59) ; therefore the moment of 

one side of the water-plane about 

the intersection is the same as the 

moment of the other side, and 

consequently the line of inter- 

section passes through the centre 

of gravity of the water-plane. 

The centre of gravity of the water- 

plane is termed the centre of flota- 

tion. In whatever direction a 

ship is inclined, transversely, 

longitudinally, or in any interme- 

diate direction, through a small 

angle, the line of intersection of 

the new water-plane with the 

original water-plane must always 

pass through the centre of flotation. For transverse inclinations 

of a ship the line of intersection is the centre line of the water- 

plane ; for longitudinal inclinations the fore-and-aft position of 

the centre of flotation has to be calculated, as we shall see 

when we deal with longitudinal inclinations. 




TOO 



Theoretical Naval Architecture. 



w 



Shift of the Centre of Gravity of a Figure due to 
the Shift of a Portion of the Figure. In Fig. 48 the 
figure PQRSTU is made up of the two portions PQTU and 

QRST, with centres of 
S gravity at g and g / respec- 

tively. Let a, d be the 
areas, the whole area a -f a' 
= A. Then the C.G. of 
the whole area is at G, such 
that a X gG = a' x /G, or 

g=J U the C.G. di- 

vides the line joining g and 
g' inversely as the areas. 
If now the portion QRST 
is shifted to the position 
UTVW with C.G. g\ the 



FIG 48. 



C.G. of the new combination PQWV is on the line^g-" at G' 
such that 

^G' = a = /G* 
Therefore by the properties of triangles GG' is parallel to g'g". 

Also 

Now, taking moments about g we have a X gg 1 = A x ^G 
. GG' = a 
"g'g" A 

or GG', the shift of the C.G. = ^ x g'g" 

or the whole area multiplied by its shift equals the small area 
multiplied by its shift, and these shifts are in parallel directions. 
Also for the horizontal shift, A X gG' = a x gg", and for the 
vertical shift, A X G^ = a X g'g. The above proof is perfectly 
general, although a simple figure has been taken by which its 
truth may be readily seen. It applies equally to the shift of 
weights. 



Conditions of Equilibrium, Transverse Metacentre, etc. IOI 

The uses that are made of this will become more apparent 
as we proceed, but the following examples will serve as illus- 
trations : 

Example. A vessel weighing W tons has a weight w tons on the deck. 
This is shifted transversely across the deck a distance of d feet, as in Fig. 49. 
Find the shift of the C.G. of the vessel both in direction and amount. 



W 



A 



FIG. 49. 

G will move to G' such that GG' will be parallel to the line joining 
the original and final positions of the weight w ; 



If w = 70 tons, d = 30 feet, W = 5000 tons, then 

GG' = 7 X3 = f& feet = 0-42 foot 
5000 

Example. In a vessel of 4000 tons displacement, suppose 100 tons ol 
coal to be shifted so that its C.G. moves 18 feet transversely and 4^ feet 
vertically. Find the shift of the C.G. of the vessel. 

The C.G. will move horizontally an amount equal to - = 0*45 ft. 
and vertically an amount equal to - = O' 1 1 ft. 



Moment of Inertia. We have dealt in Chapter II. with 



102 



Theoretical Naval Architecture. 



the moment of a force about a given point, and we denned it as 
the product of the force and the perpendicular distance of its 
line of action from the point ; also the moment of an area 
about a given axis as being the area multiplied by the distance 
of its centre of gravity from the axis. We could find the 
moment of a large area about a given axis by dividing it into 
a number of small areas and summing up the moments of all 
these small areas about the axis. In this we notice that the 
area or force is multiplied simply by the distance. Now we 
have to go a step further, and imagine that each small area is 
multiplied by the square of its distance from a given axis. If 
all such products are added together for an area, we should 
obtain not the simple moment, but what may be termed the 

01 



y 




FIG. 50. 

moment of the second degree, or more often the moment of 
inertia of the area about the given axis. 1 We therefore define 
the moment of inertia of an area about a given axis as 
follows : 

1 This is the geometrical moment of inertia. Strictly speaking, moment 
of inertia involves the mass of the body. We make here the same assump- 
tion that we did in simple moments (p. 49), viz. that the area is the 
surface of a very thin lamina or plate of homogeneous material of uniform 
thickness. 



Conditions of Equilibrium, Transverse Metacentre, etc. 103 



Imagine the. area divided into very small areas, and each such 
small area multiplied by the square of its distance from the given 
axis ; then, if all these products be added together, we shall obtain 
the moment of inertia of the total area about the given axis. 

Thus in Fig. 50, let OO be the axis. Take a very small 
area, calling it //A, distance^ from the axis. Then the sum 
of all such products as dh. x 7 2 , or (using the notation we have 
employed) // 2 . dA, will be the moment of inertia of the area 
about the axis OO. 

To determine this for any figure requires the application of 
advanced mathematics, but the result for certain regular figures 
are given below. 

It is found that we can always express the moment of 
inertia, often written I, of a plane area about a given axis by 
the expression 



where A is the area of the figure ; 

h is the depth of the figure perpendicular to the axis ; 
n is a coefficient depending on the shape of the figure 
and the position of the axis. 

First, when the axis is through the centre of gravity of 
the figure parallel to the base, as in Figs. 51 and 52 



**. 



l 







FIG 51. 

for a circle n 

for a rectangle n 

for a triangle 



FIG. 52. 

r, so that I = yV 

r, I = 
I 

1 9 B * "" " 



IO4 Theoretical Naval Architecture. 

Second, when the axis is one of the sides 

for a rectangle n = J, so that I = 
for a triangle n = -|, I = 

Example. Two squares of side a are joined to form a rectangle. The 
I of each square about the common side is 

J(a 2 )a 2 (a 1 area) 

the I of both about the common side will be the sum of each taken 
separately, or 

If, however, we took the whole figure and treated it as a rectangle, its I 
about the common side would be 

^(2fl')(2a) = \a> (area = 2a*) 

which is the same result as was obtained before. 

To find the moment of inertia of a plane figure about an axis 
parallel to and a given distance from an axis through its centre 
of gravity. 

Suppose the moment of inertia about the axis NN passing 
through the centre of gravity of the figure (Fig. 53) is I , the 



0. 





lit 




J 
fa 

J j 



JN. 



0. 



area of the figure is A, and 
OO, the given axis, is parallel 
to NN and a distance y 
from it. Then the moment 
of inertia (I) of the figure 
about OO is given by 

I = I + A/ 

The moment of inertia of an 
area about any axis is there- 
fore determined by adding 
to the moment of inertia of 
the area about a parallel axis 
through the centre of gravity, 
the product of the area into 
the square of the distance 
between the two axes. We 



FIG. S3- 

see from this that the moment of inertia of a figure about an 
axis through its own centre of gravity is always less than about 
any other axis parallel to it. 



Conditions of Equilibrium, Transverse Metacentre etc. 105 

Example. Having given the moment of inertia of the triangle in 
Fig. 52 about the axis NN through the centre of gravity as T gA/fc 2 , find the 
moment of inertia about the base parallel to NN. 

Applying the above rule, we have 

A 



which agrees with the value given above for the moment of inertia of a 
triangle about its base. 

Example. Find the moment of inertia of a triangle of area A and 
height h about an axis through the vertex parallel to the base. 

Ans. JA^ 2 . 

Example. A rectangle is 4 inches long and 3 inches broad. Compare 
the ratio of its moment of inertia about an axis through the centre parallel 
to the long and short sides respectively. 

Ans. 9 : 16. 

Example. A square of 12 inches side has another symmetrical square 
of half its area cut out of the centre. Compare the moments of inertia 
about an axis through the centre parallel to one side of, the original 
square, the square cut out, the remaining area. 

Ans. As 4 : i I 3, the ratio of the areas being 4:252. 

This last example illustrates the important fact that if an 
area is distributed away from the centre of gravity, the moment 
of inertia is very much greater than if the same area were 
massed near the centre of gravity. 

To find the Moment of Inertia of a Plane Cur- 
vilinear Figure (as Fig. 36, p. 59) about its Base. Take 
a strip PQ of length y and breadth (indefinitely small) dx> 
Then, if we regard PQ as a rectangle, its moment of inertia 
about the base DC is 



\(y . dx)y^ y 3 . dx ( y . dx = area) 

and the moment of inertia of the whole figure about DC will 
be the sum of all such expressions as this ; or 



that is, we put the third part of the cubes of the ordinates of the 
curve through either of Simpson's rules. For the water-plane 
of a ship (for which we usually require to find the moment of 
inertia about the centre line), we must add the moment of 
inertia of both sides together: and, since these are symmetrical, 
we have 

1 = I /y d* (y = semi-ordinate of water-plane). 



io6 



Theoretical Naval Architecture 



In finding the moment of inertia of a water-plane about the 
centre line, the work is arranged as follows : 



Number of 
ordinate. 


Semi-ordinates 
of water-plane. 


Cubes of 
semi-ordinates. 


Simpson's 
multipliers. 


Functions of 
cubes. 


T 


0-05 





, 





2 


4'65 


IOI 


4 


404 


3 


10-05 


1015 


2 


2,030 


4 


I4'3O 


2924 


4. 


11,696 


5 


16*75 


4699 


2 


9,398 


6 


17-65 


5498 


4 


21,992 


7 


17-40 


5268 


2 


10,536 


8 


l6'2O 


4252 


4 


17,003 


9 


13-55 


2488 


2 


4,976 


10 




899 


4 


3,596 


ii 


3-65 


49 


I 


49 



81,685 

Common interval = 28 feet 
Moment of inertia = 81,685 X f X ^ = 508,262 x 

The semi-ordinates are placed in column 2, and the cubes 
of these are placed in column 3. It is not necessary, in ordi- 
nary cases, to put any decimal places in the cube ; the nearest 
whole number is sufficient. A table of cubes is given in 
the Appendix for numbers up to 45, rising by 0*05. These 
cubes are put through Simpson's multipliers in the ordinary 
way, giving column 5. The sum of the functions of cubes 
has to be treated as follows : First there is the multiplier for 
Simpson's rule, viz. \ x 28, and then the of the expres- 
sion /_y 3 . dx> which takes into account both sides. The 
multiplier, therefore, is f X ", and the sum of the numbers 
in column 5 multiplied by this will give the moment of inertia 
required. 

Approximation to the Moment of Inertia of a 
Ship's Water-plane about the Centre Line. We have 
seen that for certain regular figures we can express the moment 
of inertia about an axis through the centre of gravity in the 
form A/fc 2 , where n is a coefficient varying for each figure. 
We can, in the same way, express the I of a water-plane area 

1 This calculation for the L.W.P. is usually done on the displacement 
sheet. Brown's displacement sheet provides a column for each water-liue 
except the two lowest. 



Conditions of Equilibrium . Transverse Metacentre, etc. \ 07 

about the centre line, but it is not convenient to use the area 
as we have done above. We know that the area can be 
expressed in the form 

k x L X B 

where L is the extreme length ; 
B breadth; 

k is a coefficient of fineness ; 

so that we can write 

I = LB 8 

where n is a new coefficient that will vary for different shapes 
of water-planes. If we can find what the values of the co- 
efficient n are for ordinary water-planes, it would be very 
useful in checking our calculation work. Taking the case of 
a L.W.P. in the form of a rectangle, we should find that n = 
0*08, and for a L.W.P. in the form of two triangles, n = 0*02. 

These are two extreme cases, and we should expect for 
ordinary ships the value of the coefficient n would lie between 
these values. This is found to be the case, and we may take 
the following approximate values for the value of n in the 
formula I = nLE 3 : 

For ships whose load water-planes are extremely fine ... 0*04 
,, ,, ,, moderately tine ... 0*05 

,, ,, ,, ,, very full ... ... O'o6 

For the water-plane whose moment of inertia we calcu- 
lated above, we have, length 280 feet, breadth 35-3 feet, and 
I = 508,262 in foot-units. Therefore the value of the coefficient 

n is 

508262 

280 x (35-3)' = ' 41 

Formula for finding the Distance of the Trans 
verse Metacentre above the Centre of Buoyancy 
(BM). We have already discussed in Chapter II. how the 
position of the centre of buoyancy can be determined if the under- 
water form of the ship is known, and now we proceed to discuss 
how the distance BM is found. Knowing this, we are able to 
fix the position of the transverse metacentre in the ship. 



io8 Theoretical Naval Architecture. 

Let Fig. 45, p. 94, represent a ship heeled over to a very 
small angle 6 (much exaggerated in the figure). 

B is the centre of buoyancy in the upright position when 

floating at the water-line WL. 
B' is the centre of buoyancy in the inclined position when 

floating at the water-line W'L'. 
v is the volume of either the immersed wedge LSL or the 

emerged wedge WSW. 
V is the total volume of displacement. 
g is the centre of gravity of the emerged wedge. 
g is the centre of gravity of the immersed wedge. 
Then, using the principle given on p. 100, BB' will be parallel 
to gg, and 



since the new displacement is formed by taking away the wedge 
WSW from the original displacement, and putting it in the 
position LSL'. 

Now for the very small angle of inclination, we may say 
tliat 

BB' 

BM = sln * 
or BB' = BM sin 

so that we can find BM if we can determine the value of 
v x gg t since V, the volume of displacement, is known. 

Let Fig. 54 be a section of the vessel; //, //', the original 
and new water-lines respectively, the angle of inclination being 
very small. Then we may term wSu/ the emerged triangle, 
and /S/ the immersed triangle, being transverse sections of 
the emerged and immersed wedges, and a/a/, // being for all 
practical purposes straight lines. If y be the half-breadth of 
the water-line at this section, we can say ww' = II' = y sin 0, 
and the area of either of the triangles is 

\y X y sin = ^y z sin 6 

Let 0, d be the centres of gravity of the triangles #>Sze/, /S/ 
respectively ; then we can say, seeing that 6 is very small, that 



Conditions of Equilibrium, Transverse Metacentre, etc. 109 

ad = f.y, since the centre of gravity of a triangle is two-thirds 
the height from the apex. The new immersed section being 
regarded as formed by the transference of the triangle 




FIG. 



to the position occupied by the triangle /S/, the moment of 
transference is 

(l/ sin 0) X & = f / sin 



and for a very small length dx of the water-line the moment 
will be 

f y sin . dx 

since the small volume is \f sin . dx, and the shift of its 
centre of gravity is f jy. If now we summed all such expres- 
sions as this for the whole length of the ship, we should get 
the moment of the transference of the wedge, or v x gg'. 
Therefore we may say, using the ordinary notation 

vxgg' = /fy sin e . dx 
= fsin0jy.dk 

therefore we have 

v X gg __ | sin jy . dx 

~V ~~ " v 



BB' = BM sin = 



or BM = 



But the numerator of this expression is what we have found to 



no Theoretical Naval Architecture. 

be the moment of inertia of 'a water-plane about its centre line, 
y being a semi-ordinate ; therefore we can write 



We have seen, on p. 105, how the moment of inertia of a 
water-plane is found for any given case, and knowing the 
volume of displacement, we can then determine the distance 
BM, and so, knowing the position of the C.B., fix the position 
of the transverse metacentre in the ship. 

Example. A lighter is in the form of a box, 120 feet long, 30 feet 
broad, and floats at a draught of 10 feet. Find its transverse BM. 

In this case the water-plane is a rectangle 120' X 30', and we want its 
I about the middle line. Using the formula for the I of a rectangle about 
an axis through its centre parallel to a side, nA/4 2 , we have 

1 = n X 3 6 oo X 900 (k = 30) 

= 270,000 

V, the volume of displacement, = 120x30x10 = 36,000 
BM = 270,000 5feet 

36,000 

Example. A pontoon of 10 feet draught has a constant sectio*n in the 
form of a trapezoid, breadth at the water-line 30 feet, breadth at base 
20 feet, length i'2O feet. Find the transverse BM. 

Ans. 9 feet. 

It will be noticed that the water-plane in this question is 
the same as in the previous question, but the displacement being 
less, the BM is greater. M is therefore higher in the ship for 
two reasons. BM is greater and B is higher in the second case. 

Example. A raft is formed of two cylinders 5 feet in diameter, parallel 
throughout their lengths, and 10 feet apart, centre to centre. The raft floats 
with the axes of the cylinders in the surface. Find the transverse BM. 

We shall find that the length does not affect the result, but we will 
suppose the length is / feet. We may find the I of the water-plane in two 
ways. It consists of two rectangles each /' X 5', and their centre lines 
are 10 feet apart. 

1. The water -plane may be regarded as formed by cutting a rectangle 
/' x 5' out of a rectangle /' X 15' ; 

.-. I - (/ x 15) x is 2 - A(/ x 5) x s 3 

= Mi5 8 - 5 3 ) 
= af|V 

this being about a fore-and-aft axis at the centre of the raft. 

2. We may take the two rectangles separately, and find the I of each 
about the centre line of the raft, which is 5 feet from the line through the 
centre of each rectangle. Using the formula 

I = I. + Ay' 

|(/ x 5)5* + (/ x 5)5* 



Conditions of Equilibrium^ Transverse Metacentre, etc. 1 1 1 

and for both rectangles the moment of inertia will be twice this, or 3 f| 8 /, as 
obtained above. 

We have to find the volume of displacement, which works out to *$l 
cubic feet. The distance BM is therefore 

s$o/-4-^./= 1 3- 8 feet 

Example. A raft is formed of three cylinders, 5 feet in diameter, 
parallel and symmetrical throughout their lengths, the breadth extreme 
being 25 feet. The raft floats with the axes of the cylinders in the surface. 
Find the transverse BM. 

The moment of inertia of the water-plane of this raft is best found by 
using the formula I = I + Ay 8 for the two outside rectangles, and adding 
it to I , the moment of inertia of the centre rectangle about the middle line. 
We therefore have for the whole water-plane I = <-\p/, where / = the 
length ; and the volume of displacement being - 8 2 2 8 5 /, the value of BM will be 
35 ^et. 

Approximate Formula for the Height of the Trans- 
verse Metacentre above the Centre of Buoyancy. 
The formula for BM is 



We have seen that we may express I as LB 8 , where n is 
a coefficient which varies for different shapes of water-planes, 
but which will be the same for two ships whose water-planes 
are similar. 

We have also seen that we may express V as LBD, where 
D is the mean moulded draft (to top of keel amidships), and k 
is a coefficient which varies for different forms, but which will 
be the same for two ships whose under- water forms are similar. 
Therefore we may say 

n X L X B 3 



BM = 



B 2 



where a is a coefficient obtained from the coefficients n and k. 
Sir William White, in the "Manual of Naval Architecture," 
gives the value of a as being between 0*08 and 0*1, a usual 
value for merchant ships being 0*09. The above formula 
shows very clearly that the breadth is more effective than the 
draught in determining what the value of BM is in any given 
case. It will also be noticed that the length is not brought in. 



112 



Theoretical Naval Architecture, 



The ship for which the moment of inertia of a water-plane 
was calculated on p. 106, had a displacement of 1837 tons up 
to that water-plane. The value of BM is therefore 



508262 
X 35 



7*91 feet 



The breadth and mean draught were 35*3 and 13! feet re- 
spectively. Consequently the value of the coefficient a is 
0*084. 

To prove that a Homogeneous Log of Timber of 
Square Section and Specific Gravity P 5 cannot float 
in Fresh Water with One of its Faces Horizontal. 
The log having a specific gravity of 0-5 will float, and will float 
with half its substance immersed. The condition that it shall 
float in stable equilibrium, as regards transverse inclination, in 
any position is that the transverse metacentre shall be above 
the centre of gravity. 

Let the section be as indicated in Fig. 55, with side length 

20. And suppose the log 
is placed in the water with 
one side of this section 
horizontal. Then the 
draught-line will be at a 
distance a from the bot- 
tom, and the log, being 
homogeneous, i.e. of the 
same quality all through, 
will have its C.G. in the 
middle at G, at a distance 
also of a from the bottom. 
The centre of buoyancy 
FIG. 55. will be at a distance of 

- from the bottom. The height of the transverse metacentre 

2 

above the centre of buoyancy is given by 



! I 

j 42a,- 









BM 



I 
V 



Conditions of Equilibrium, Transverse Metacentre, etc. 113 

where I = moment of inertia of water-plane about a longi- 

tudinal axis through its centre 
V = volume of displacement in cubic feet. 

Now, the water-plane of the log is a rectangle of length / 
and breadth 20, and therefore 



its I = j /. za(2of 
and V = /. 2a . a = zla? 
:. BM = iV^ 3 -r- 2la* = a 
But BG = \a 

therefore the transverse metacentre is below the centre of 
gravity, and consequently the log cannot float in the position 
given. 

If, now, the log be assumed floating with one corner down- 
ward, it will be found by a precisely similar method that 

BG = 0-4710 
and BM = 0*9430 

Thus in this case the transverse metacentre is above the 
centre of gravity, and consequently the log will float in stable 
equilibrium. 

It can also be shown by similar methods that the position 
of stable equilibrium for all directions of inclination of a cube 
composed of homogeneous material of specific gravity o'5 is 
with one corner downwards. 

Metacentric Diagram. We have seen how the position 
of the transverse metacentre can be determined for any given 
ship floating at a definite water-line. It is often necessary, 
however, to know the position of the metacentre when the ship 
is floating at some different water-line ; as, for instance, when 
coal or stores have been consumed, or when the ship is in a light 
condition. It is usual to construct a diagram which will show 
at once, for any given mean draught which the vessel may have, 
the position of the transverse metacentre. Such a diagram is 
shown in Fig. 56, and it is constructed in the following manner: 
A line W^ is drawn to represent the load water-line, and 
parallel to it are drawn W 2 L 2 , W 3 L 2 , W 4 L 4 to represent the 

i 



Theoretical Naval Architecture. 



water-lines Nos. 2, 3, and 4, which are used for calculating the 
displacement, the proper distance apart, a convenient scale 
being \ inch to i foot. A line L^ 4 is drawn cutting these 
level lines, and inclined to them at an angle of 45. Through 
the points of intersection L 1} L 2 , L 3 , L 4 , are drawn vertical lines 
as shown. The ship is then supposed to float successively at 
these water-lines, and the position of the centre of buoyancy 
and the distance of the transverse metacentre above the C.B. 



il 



W, 



w 





FIG. 56. 



calculated for each case. The methods employed for finding 
the position of the C.B. at the different water-lines have already 
been dealt with in Chapter II. On the vertical lines are then 
set down from the L.W.L. the respective distances of the 
centres of buoyancy below the L.W.L. Thus L^ is the 
distance when floating at the L.W.L., and AB 3 the distance 
when floating at No. 3 W.L. In this way the points B 15 B 2 , 
B 3 , B 4 are obtained ; and if the calculations are correct, a fair 



Conditions of Equilibrium, Transverse Metacentre, etc. 115 

line can be drawn passing through all these spots as shown. 
Such a curve is termed the curve of centres of buoyancy. It is 
usually found to be rather a flat curve, being straight near the 
load-line condition. The distance BM for each water-line is 
then set up from Bi, B 2 , B 3 , B 4 respectively, giving the points 
MU M 2 , M 3 , M 4 . A curve can then be drawn through these 
points, which is termed the curve of transverse metacentres. 
Now, suppose the ship is floating at some intermediate water- 
line say wl: through /, where wl cuts the 45 line, draw a 
vertical cutting the curves of centres of buoyancy and meta- 
centres in ]b and m respectively. Then m will be the position 
of the transverse metacentre of the ship when floating at the 
water-line wl. It will be noticed that we have supposed the 
ship to float always with the water-plane parallel to the L.W.P.; 
that is to say, she does not alter trim. For water-planes not 
parallel to the L.W.P. we take the mean draught (i.e. the 
draughts at the fore-and-aft perpendiculars are added together 
and divided by 2), and find the position of M on the meta- 
centric diagram for the water-plane, parallel to the L.W.P., 
corresponding to this mean draught. Unless the change of 
trim is very considerable, this is found to be correct enough 
for all practical purposes. Suppose, however, the ship trims 
very much by the stern, 1 owing to coal or stores forward being 
consumed, the shape of her water-plane will be very different 
from the shape it would have if she were floating at her normal 
trim or parallel to the L.W.P. ; generally the water-plane will 
be fuller under these circumstances, and the moment of inertia 
will be greater, and consequently M higher in the ship, than 
would be given on the metacentric diagram. When a ship 
is inclined, an operation that will be described later, she 
is frequently in an unfinished condition, and trims consider- 
ably by the stern. It is necessary to know the position of 
the transverse metacentre accurately for this condition, and 

1 This would be the case in the following : A ship is designed to float 
at a draught of 17 feet forward and 19 feet aft, or, as we say, 2 feet by the 
stern. If her draught is, say, 16 feet forward and 20 feet aft, she will have 
the same mean draught as designed, vk. 18 feet, but she will trim 2 feet 
more by the stern. 



1 1 6 Theoretical Naval A rchitecture. 

consequently the metacentric diagram cannot be used, but a 
separate calculation made for the water-plane at which the 
vessel is floating. 

On the metacentric diagram is placed also the position of 
the centre of gravity of the ship under certain conditions. For 
a merchant ship these conditions may vary considerably owing 
to the nature of the cargo carried. There are two conditions 
for which the C.G. may be readily determined, viz. the light 
condition, and the condition when loaded to the load-line with 
a homogeneous cargo. The light condition may be denned as 
follows : No cargo, coal, stores, or any weights on board not 
actually forming a part of the hull and machinery, but includ- 
ing the water in boilers and condensers. The draught-lines for 
the various conditions are put on the metacentric diagram, and 
the position of the centre of gravity for each condition placed 
in its proper vertical position. The various values for GM, the 
metacentric height, are thus obtained. 

On the left of the diagram are placed, at the various water- 
lines, the mean draught, displacement, and tons per inch. 1 

There are two forms of section for which it is instructive to 
construct the metacentric diagram. 

1. A floating body of constant rectangular section. 

2. A floating body of constant triangular section, the apex 
of the triangle being at the bottom. 

i. For a body having a constant rectangular section, the 
moment of inertia of the water-plane is the same for all 
draughts, but the volume of displacement varies. Suppose the 
rectangular box is 80 feet long, 8 feet broad, 9 feet deep. Then 
the moment of inertia of the water-plane for all draughts is 
5*5(80 X 8) x 8" = ^4P 

The volumes of displacement are as follows : 

Draught 6 inches V = 80 X 8 X J cubic feet 

1 foot V = 8ox8 

2 feet V = 80x8x2 

4 V = Sox 8x4 .. 

V = 80x8x7 

" t 9 V =8ox 8x9 

1 For a specimen metacentric diagram, see Example 40, Chap. III. 
Specimen diagrams for various types of ships are given in the Author's 
"War Ships." 



Conditions of Equilibrium, Transverse Metacentre, etc. 1 1 7 



and the values of BM are therefore as follows : 

Draught 6 inches BM = 10*66 feet 

1 foot ... ' BM = 5-33 

2 feet BM = 2'66 

4 BM= 1-33 

7 ,, BM = 076 

9 BM = 0-59 

The centre of buoyancy is always at half-draught, so that 
its locus or path will be a straight line, 1 and if the values obtained 
above are set off from the centres of buoyancy at the various 
water-lines, we shall obtain the curve of transverse metacentres 
as shown in Fig. 57 by the curve A A, the line BB being the 
corresponding locus of the centres of buoyancy. 



9-0. 



6-O. 



3-5. 



Ox' 



FIG. 57- 

s. For a floating body with a constant triangular section, the 
locus of centres of buoyancy is also a straight line because it is 
always two-thirds the draught above the base. 1 Suppose the 
triangular section to be 10 feet broad at the top and 9 feet deep, 
the length of the body being 120 feet. In this case we must 
calculate the moment of inertia of each water-plane and the 
volume of displacement up to each. The results are found to 
be as follows : 

' This may be seen by finding a few spots on this locus. 



uS Theoretical Naval Architecture. 

Draught I foot BM = 0*20 feet 

,, 2 feet BM = 0*41 ,, 

4 BM = o-82 

6 BM=i-2 3 

9 BM = i-85 

These values are set up from the respective centres of 
buoyancy, and give the locus of transverse metacentres, which 
is found to be a straight line, as shown by CC in Fig. 57, DD 
being the locus of centres of buoyancy. 

Approximation to Locus of Centres of Buoyancy 
on the Metacentric Diagram. We have seen (p. 65) how 
the distance of the centre of buoyancy below the L.W.L. can 
be approximately determined. The locus of centres of buoyancy 
in the metacentric diagram is, in most cases, very nearly straight 
for the portion near the load-line, and if we could obtain easily 
the direction the curve takes on leaving the position for the load 
water-line, we should obtain a very close approximation to the 
actual curve itself. It might be desirable to obtain such an 
approximation in the early stages of a design, when it would 
not be convenient to calculate the actual positions of the centre 
of buoyancy, in order to accurately construct the curve. 

Let be the angle the tangent to the curve of buoyancy at 
the load condition makes with the horizontal, as in 

Fig. 56; 

A, the area of the load water-plane in square feet ; 
V, the volume of displacement up to the load water-line 

in cubic feet ; 
^, the distance of the centre of buoyancy of the load 

displacement below the load water-line in feet. 
Then the direction of the tangent to the curve of buoyancy is 
given by 

tan 9 = -* (for proof see later.) 

Each of the terms in the latter expression are known or can 
be readily approximated to, 1 and we can thus determine the in- 
clination at which the curve of centres of buoyancy will start, 
and this will closely follow the actual curve. 2 

1 See Example 39, p. 143, for a further approximation. 
9 See a paper by the late Professor Jenkins read before the Institution 
of Naval Architects in 1884. 



Conditions of Equilibrium, Transverse Metacentre, etc. \ 19 
In a given case 

A = 7854 square feet 

h = 5*45 f e et 

V = 2140 x 35 cubic feet 

so that 



2140 x 35 i 

= 0-572* 

Finding the Metacentric Height by Experiment. 
Inclining Experiment. We have been dealing up to the 
present with the purely geometrical aspect of initial stability, 
viz. the methods employed and the principles involved in 
finding the position of the transverse metacentre. All that is 
needed in order to determine this point is the form of the 
underwater portion of the vessel. But in order to know any- 
thing about the vessel's initial stability, we must also know the 
vertical position of the centre of gravity of the ship, and it is to 
determine this point that the inclining experiment is performed. 
This is done as the vessel approaches completion, when 
weights that have yet to go on board can be determined 
together with their final positions. Weights are shifted trans- 
versely across the deck, and by using the principle explained on 
p. 100, we can tell at once the horizontal shift of the centre of 
gravity of the ship herself due to this shift of the weights on 
board. The weight of the ship can be determined by calculating 
the displacement up to the water-line she floats at, during the 
experiment. (An approximate method of determining this 
displacement when the vessel floats out of her designed trim 

1 The best way to set off this line is to set off a horizontal line of 10 feet 
long (on a convenient scale), and from the end set down a vertical line 
572 feet long on the same scale. This will give the inclination required, 

for tan = f^- = ^ = 0-572. 

base 10 

This remark applies to any case in which an angle has to be set off 
very accurately. A table of tangents is consulted and the tangent of 
the required angle is found, and a similar process to the above is gone 
through. 



I2O 



Theoretical Naval Architecture. 



will be found on p. 152.) Using the notation employed on 
p. 100, and illustrated by Fig. 49, we have 

rr , wxd 
GG = IT" 

Now, unless prevented by external forces, it is evident that 
the vessel must incline over to such an angle that the centre of 
gravity G' and the centre of buoyancy B' are in the same verti- 
cal line (see Fig. 58), and, the angle of inclination being small, 







FIG. 58. 

M will be the transverse metacentre. If now we call 6 the 
angle of inclination to the upright, GM being the " metacentric 
height " 

GG' 



w X d 



~ W X tan 

using the value found above for GG'. The only term that we 
do not yet know in this expression is tan 6, and this is found in 
the following manner : At two or three convenient positions 



Conditions of Equilibrium, Transverse Metacentre, etc. 1 2 1 

in the ship x (such as at bulkheads or down hatchways) plumb- 
bobs are suspended from a point in the middle line of the ship, 
and at a convenient distance from the point of suspension a 
horizontal batten is fixed, with the centre line of the ship marked 
on it, as shown by PQ in Fig. 58. Before the ship is inclined, 
the plumb-line should coincide, as nearly as possible, with the 
centre-line of the ship that is to say, the ship should be prac- 
tically upright. When the ship is heeled over to the angle 0, 
the plumb-line will also be inclined at the same angle, 0, to the 
original vertical or centre line of the ship, and if / be the 
distance of the horizontal batten below the point of suspension 
O in inches, and a the deviation of the plumb-line along the 
batten, also in inches, the angle 6 is at once determined, for 

tan 6 = -, 
so that we can write l 




In practice it is convenient to check the results obtained, by 
dividing the weight w into four equal parts, placing two sets on 
one side and two sets on the other side, arranged as in Fig. 59. 

The experiment is then performed in the following order : 

(a) See if the ship is floating upright, in which case the 
plumb-lines will coincide with the centre of the ship. 

(b) The weight (i), Fig. 59, is shifted from port to star- 
board on to the top of weight (3) through the distance d feet, 
say, and the deviations of the plumb-lines are noted when the 
ship settles down at a steady angle. 

(c) The weight (2) is shifted from port to starboard on to 
the top of weight (4) through the distance d feet, and the 
deviations of the plumb-line noted. 

(d) The weights (i) and (2) are replaced in their original posi- 
tions, when the vessel should again resume her upright position. 

* If two positions are taken, one is forward and the other aft. If three 
positions are taken, one is forward, one aft, and one amidships. 

* This depends on the assumption that M is a fixed point for the heel 
obtained, and this is true for ordinary ships. It fails, however, in the case 
of a vessel of very small or of zero metacentric height. See examples in 
Appendix A. 



122 Theoretical Naval Architecture. 

(e) The weight (3) is moved from starboard to port, and 
the deviations of the plumb-lines noted. 

(/) The weight (4) is moved from starboard to port, and 
the deviations of the plumb-lines noted. 

With the above method of conducting the experiment, 1 and 
using two plumb-lines, we obtain eight readings, and if three 
plumb-lines were used we should obtain twelve readings. It is 
important that such checks should be obtained, as a single result 
might be rendered quite incorrect, owing to the influence of the 
hawsers, etc. A specimen experiment is given on p. 123, in 
which two plumb-lines were used. The deviations obtained 



! 


a. j 


_r* ~H 
mm 


mnh 


i 

1 
1 


! 



FIG. 59. 

are set out in detail, the mean deviation for a shift of 12^ tons 
through 36 feet being 5 inches, or the mean deviation for a 
shift of 25 tons through 36 feet is 10^ inches. 

Precautions to be taken when performing an Inclining Experi- 
ment. A rough estimate should be made of the GM expected 
at the time of the experiment \ the weight of ballast can then be 
determined which will give an inclination of about 4 or 5 when 
one-half is moved a known distance across the deck. The weight 
of ballast thus found can then be got ready for the experiment. 

A personal inspection should be made to see that all weights 
likely to shift are efficiently secured, the ship cleared of all 

1 There is a slight rise of G, the centre of gravity of the ship, in this 
method : but the error involved is inappreciable. 



Conditions of Equilibrium, Transverse Metacentre, etc. 123 

free water, and boilers either emptied or run up quite full. 
Any floating stages should be released or secured by veiy slack 
painters. 

If possible a fine day should be chosen, with the water calm 
and little wind. All men not actually employed on the experi- 
ment should be sent ashore. Saturday afternoon or a dinner 
hour is found a convenient time, since then the majority of 
the workmen employed finishing the ship are likely to be away. 

The ship should be hauled head or stern on to the wind, 
if any, and secured by hawsers at the bow and stern. When 
taking the readings, these hawsers should be slacked out, so as 
to ensure that they do not influence the reading. The ship 
should be plumbed upright before commencing. 

An account should be taken, with positions of all weights to 
be placed on board to complete, of all weights to be removed, 
such as yard plant, etc., and all weights that have to be shifted. 

The following is a specimen report of an inclining experi- 
ment : 



Report on Inclining Experiment performed on " 



at . Density of water cubic feet to the ton. 



-, 189-, 



Draught of water 

i j 

Displacement in tons at this draught 



1 6' 9" forward. 
22' 10" aft. 
5372 



The wind was slight, and the ship was kept head to wind during the 
experiment. Ballast used for inclining, 50 tons. Lengths of pendulums, 
two in number, 15 feet. Shift of ballast across deck, 36 feet. 





Deviation of pendulum in 15 feet. 


Forward. 


Aft. 


Experiment I, 12^ tons port to starboard 

>i 2, I2j ,, ,, 

Ballast replaced, zero checked ... 
Experiment 3, 12^ tons starboard to port 
4, I2i 


5g 

10*" 

right 

$ 


5*" 

io|" 
right 

s; 

I0j" 



The condition of the ship at the time of inclining is as defined below : 

Bilges dry. 

Water -tanks empty. 






124 Theoretical Naval Architecture. 

No water in boilers, feed-tanks, condensers, distillers, cisterns, etc. 

Workmen on board, 66. 

Tools on board, 5 tons. 

Masts and spars complete. 

No boats on board. 

Bunkers full. 

Anchors and cables, complete and stowed. 

No provisions or stores on board. 

Engineers' stores, half on board. 

Hull complete. 

The mean deviation in 15 feet for a shift of 25 tons through 36 feet is 
lOfk inches = 10*312 inches. 

... GM = 2 5 x 3 6 x 15 X 12 = 
10-312 x 5372 

The ship being in an incomplete condition at the time of 
the inclining experiment, it was necessary to take an accurate 
account of all weights that had to go on board to complete, 
with their positions in the ship, together with an account of 
all weights that had to be removed, with their positions. The 
total weights were then obtained, together with the position of 
their final centre of gravity, both in a longitudinal and vertical 
direction. For the ship of which the inclining experiment is 
given above, it was found that to fully complete her a total 
weight of 595 tons had to be placed on board, having its 
centre of gravity u feet before the midship ordinate, and 3*05 
feet below the designed L.W.L. Also 63 tons of yard plant, 
men, etc., had to be removed, with centre of gravity 14 feet 
abaft the midship ordinate, and 15 feet above the designed 
L.W.L. The centre of buoyancy of the ship at the experi- 
mental water-line was 10*8 feet abaft the midship ordinate, 
and the transverse metacentre at this line was calculated at 
3-14 feet above the designed L.W.L. 

We may now calculate the final position of the centre of 
gravity of the completed ship as follows, remembering that 
in the experimental condition the centre of gravity must be 
in the same vertical line as the centre of buoyancy. The 
vertical position of G in the experimental condition is found 
by subtracting the experimental GM, viz. 2*92 feet, from the 
height of the metacentre above the L.W L. as given above, 
viz. 3' 1 4 feet. 



Conditions of Equilibrium, Transverse Metacentt <?, etc. 125 





H 


Above 
L.W.L. 


Below 
L.W.L. 


Abaft 
amidships. 


Bel 
amids 

C 
j 


ore 
hips. 


5 


Moment. 


3^5 


Moment. 


i 

% 


Moment. 


Moment. 


Weight of ship at time"! 
of experiment ...J 
Weight to go on board) 
to complete ... .../ 

Weight to be takenj 
from ship / 


5372 

595 


0'22 


Il82 


1813 


10-8 


58,017 


II 


6545 


5967 
63 


15 


Il82 

945 





1813 


14-0 


58,017 
882 


6545 


5904 237 1813 57,135 6545 
237 6,545 



1576 



50,590 



The final position of the centre of gravity of the ship is 
therefore 

= 0-266 feet below the L.W.L. 
= 8 '57 feet abaft amidships 



the final displacement being 5904 tons. 

The mean draught corresponding to the displacement can 
be found by the methods we have already dealt with, and corre- 
sponding to this draught, we can find on the metacentric 
diagram the position of the transverse metacentre. In this case 
the metacentre was 273 feet above the L.W.L., and conse- 
quently the value of GM for the completed condition was 

273 + 0^266 = 2^996 feet 

or say, for all practical purposes, that the transverse metacentric 
height in the completed condition was 3 feet. 

It is also possible to ascertain what the draughts forward 
and aft will be in the completed condition, as we shall see in 
the next chapter. 

Values of GM, the "Metacentric Height." We 
have discussed in this chapter the methods adopted to find 
for a given ship the value of the transverse metacentric height 
GM. This distance depends upon two things : the position of 
G, the centre of gravity of the ship and the position of M, the 



126 Theoretical Naval Architecture. 

transverse melacentre. The first is dependent on the vertical 
distribution of the weights forming the structure and lading 01 
the ship, and its position in the ship must vary with differences 
in the disposition of the cargo carried. The transverse meta- 
centre depends solely upon the form of the ship, and its 
position can be completely determined for any given draught 
of water when we have the sheer drawing of the vessel. There 
are two steps to be taken in finding its position for any given 
ship floating at a certain water-line. 

1. We must find the vertical position of the centre of 
buoyancy, the methods adopted being explained in Chapter II. 

2. We then find the distance separating the centre of 
buoyancy and the transverse metacentre, or BM, as explained 
in the present chapter. 

By this means we determine the position of M in the ship. 

The methods of estimating the position of G, the centre 
of gravity for a new ship, will be dealt with separately in 
Chapter VI. ; but we have already seen how the position of G 
can be determined for a given ship by means of the inclining 
experiment. Having thus obtained the position of M and G in 
the ship, we get the distance GM, or the metacentric height. 

The following table gives the values of the metacentric height 
in certain classes of ships. For fuller information reference 
must be made to the works quoted at the end of the book. 



Type of -ship. 


Values of GM. 


Harbour vessels, as tugs, etc 
Modern protected cruisers... 
Modern British battleships 
Older central citadel armourclads ... 
Shallow-draught gunboats for river service 
Merchant steamers (varying according to 1 
the nature and distribution of the cargo) / 


15 to 18 inches 
2 to 2j feet 
4 to 5 feet 
4 to 8 feet 
12 feet 

I to 3 feet 
1 to 3i feet 







The amount of metacentric height given to a vessel is based 
largely upon experience with successful ships. In order that 
a vessel may be " stiff" that is, difficult to incline by external 
forces as, for example, by the pressure of the wind on the 



Conditions of Equilibrium, Transverse Metacentre, etc. 1 27 

sails the metacentric height must be large. This is seen by 
reference to the expression for the moment of statical stability 
at small angles of inclination from the upright, viz. ^i#'-' 

W X GM sin (see p. 98) 

W being the weight of the ship in tons ; being the angle of 
inclination, supposed small. This, being the moment tending 
to right the ship, is directly dependent on GM. A " crank " 
ship is a ship very easily inclined, and in such a ship the 
metacentric height is small. For steadiness in a seaway the 
metacentric height must be small. 

There are thus two opposing conditions to fulfil 

j. The metacentric height GM must be enough to enable 
the ship to resist inclination by external forces. This is espe- 
cially the case in sailing-ships, in order that they may be able 
to stand up under canvas without heeling too much. In the 
case of the older battleships with short armour belts and 
unprotected ends, sufficient metacentric height had to be pro- 
vided to allow of the ends being riddled, and the consequent 
reduction of the moment of inertia of the water-plane. 

2. The metacentric height must be moderate enough (if 
this can be done consistently with other conditions being 
satisfied) to make the vessel steady in a seaway. A ship which 
has a very large GM comes back to the upright very suddenly 
after being inclined, and consequently a vessel with small 
GM is much more comfortable at sea, and, in the case of a 
man-of-war, affords a much steadier gun platform. 

In the case of sailing-ships, a metacentric height of from 
3 to 3^- feet is provided under ordinary conditions of service, 
in order to allow the vessel to stand up under her canvas. It is, 
however, quite possible that, when loaded with homogeneous 
cargoes, as wool, etc., this amount cannot be obtained, on 
account of the centre of gravity of the cargo being higl) up in 
the ship. In this case, it would be advisable to take in water 
or other ballast in order to lower the centre of gravity, and 
thus increase the metacentric height 

In merchant steamers the conditions continually vary on 
account of the varying nature and distribution of the cargo 



128 Theoretical Naval Architecture. 

carried, and it is probable that a GM of i foot should be the 
minimum provided when carrying a homogeneous cargo (con- 
sistently with satisfactory stability being obtained at large 
inclinations). 1 There are, however, cases on record of vessels 
going long voyages with a metacentric height of less than i 
foot, and being reported as comfortable and seaworthy. Mr. 
Denny (Transactions of the Institution of Naval Architects ', 
1896) mentioned a case of a merchant steamer, 320 feet long 
^'carrying a homogeneous cargo), which sailed habitually with 
a metacentric height of 0*6 of a foot, the captain reporting her 
behaviour as admirable in a seaway, and in every way com- 
fortable and safe. 

It is the practice of one large steamship company to lay 
down that the metacentric height in the loaded condition is no 
greater than is required to secure that the metacentric height 
in the light condition is not negative. 

Effect on Initial Stability due to the Presence of 
Free Water in a Ship. On reference to p. 123, where the 
inclining experiment for obtaining the vertical position of the 
centre of gravity of a ship is explained, it will be noticed that 
special attention is drawn to the necessity for ascertaining 
that no free water is allowed to remain in the ship while the 
experiment is being performed. By free water is meant water 
having a free surface. In the case of the boilers, for instance, 
they should either be emptied or run up quite full. We now 
proceed to ascertain the necessity for taking this precaution. 
If a compartment, such as a ballast tank in the double bottom, 
or a boiler, is run up quite full, it is evident that the water will 
have precisely the same effect on the ship as if it were a solid 
body having the same weight and position of its centre of 
gravity as the water, and this can be allowed for with very 
little difficulty. Suppose, however, that we have on board in 
a compartment, such as a ballast tank in the double bottom, a 
quantity of water, and the water does not completely fill the 

1 Mr. Pescod, before the North-East Coast Institution of Engineers and 
Shipbuilders, 1903, dealt with the minimum GM for small vessels. He 
there states that it is generally recognized that the GM of cargo vessels 
should not be less than O'8 foot provided that a righting arm of like amount 
is obtained at 30 to 40 degrees. 



Conditions of Equilibrium, Transverse Metacentre^ etc. 129 

tank, but has a free surface, as wl^ Fig. 60.! If the ship is 
heeled over to a small angle 0, the water in the tank must 
adjust itself so that its surface w'l' is parallel to the level water- 
line W'L'. Let the volume of either of the small wedges wsw', 
1st be z> > and g> g 1 the positions of the ircentres of gravity, b, b 1 
being the centres of gravity of the whole volume of water in 
the upright and inclined positions respectively. Then, if V 
be the total volume of water in the tank, we have 

V X bV = v Q X gg' 
and bb' = =^- X XS* 



and bit is parallel to gg. 
found the moment of 
transference of the 
wedges WSW, LSL', 
in Fig. 45, we can find 
the moment of trans- 
ference of the small 
wedges wsw', /j/,viz. 

^o X gg' = i X B 
where i is the moment 
of inertia of the free 
surface of the water in 
the tank about a fore- 
and-aft axis through s ; 
and 9 is the circular 
measure of the angle 
of inclination. 



Now, in precisely the same way as we 




FIG. 60. 



Substituting this value for v x gg', we have 

i X 



Draw the new vertical through ', meeting the middle line in 
m\ then 

bb' = bmxQ 



1 Fig. 60 is drawn out of proportion for the sake of clearness. 



J 3 Theoretical Naval Architecture. 

and consequently 

bm X = 



and bm = ^r- 



Now, if the water were solid its centre of gravity would be 
at b both in the upright and inclined conditions, but the weight 
of the water now acts through the point b' in the line b'm, and 
its effect on the ship is just the same as if it were a solid 
weight concentrated at the point m. So that, although b is 
the actual centre of gravity of the water, its effect on the ship, 
when inclined through ever so small an angle, is the same as 
though it were at the point m t and in consequence of this the 
point m is termed the virtual centre of gravity of the water. 1 
This may be made clearer by the following illustrations : 

1. Suppose that one instant the water is solid, with its 
centre of gravity at b, and the following instant it became liquid. 
Then, for small angles of inclination, its effect on the ship would 
be the same as if we had raised its weight through a vertical 
distance bm from its actual to its virtual centre of gravity. 

2. Imagine a pendulum suspended at m t with its bob at b. 
On the ship being inclined to the small angle 0,the pendulum 
will take up the position mb\ and this corresponds exactly to 
the action of the water. 

We thus see that the centre of gravity of the ship cannot be 
regarded as being at G, but as having risen to GO, and if W be 

the weight of water in tons = -f (the water being supposed 

salt), we have 

W X GG = W X bm 



and therefore 
GG = V Xm = x bm ( v = volume of displacement) 



1 See a paper by Mr. W. Hok, at the Institution of Naval Architects, 
1895, on " The Transverse Stability of Floating Vessels containing Liquids, 
with Special Reference to Ships carrying Oil in Bulk." See also a paper 
in the "Transactions of the Institution of Engineers and Shipbuilders in 
Scotland for 1889," by the late Professor Jenkins, on the stability of vessels 
carrying oil in bulk. 



Conditions of Equilibrium, Transverse Metacentre, etc. 131 

But we have seen that 

bm = ==- 
and therefore 

GG, = Y X ^ = 1 

The new moment of stability at the angle is 
W X G M x sin B = W x (GM - GG ) sin 6 

= WX 

the metacentric height being reduced by the simple expres- 
sion ==. We notice here that the amount of water does not 

affect the result, but only the moment of inertia of the free 
surface. The necessity for the precaution of clearing all free 
water out of a ship on inclining is now apparent. A small 
quantity of water will have as much effect on the position of 
the centre of gravity, and therefore on the trustworthiness of 
the result obtained, as a large quantity of water, provided it 
has the same form of free surface. If a small quantity of 
water has a large free surface, it will have more effect than 
a very large quantity of water having a smaller free surface. 
If the liquid contained is other than the water the vessel is 

P* i 
floating in, the loss of metacentric height is -^, where p is the 

specific gravity of liquid compared with outside water, and V 
the total volume of displacement. 

Example. A vessel has a compartment of the double bottom at the 
middle line, 60 feet long and 30 feet broad, partially filled with salt water. 
The total displacement is 9100 tons, and centre of gravity of the ship and 
water is o - 26 feet below the water-line. Find the loss of metacentric 
height due to the water having a free surface. 

We have here given the position of the centre of gravity of the ship and 
the water. The rise of this centre of gravity due to the mobility of the 
water is, using the above notation 

i 

V 
and / = ^(60 X 30) X (30)* 

= 5 X (30)' 

Since the free surface is a rectangle 60 feet long and 30 broad 
and V = 9100 x 35 cubic feet 

therefore the loss in metacentric height = =- = 0-424 feet 

9100 x 35 



132 Theoretical Naval Architecture. 

Met acentric Diagrams for Simple Figures. i. A 
rectangular box. This is dealt with on p. 116. 

B 2 

BM = Ta "n B = breadth D = draught, 

D B a 

and M from base is -- h ^ jj 

By the methods of the calculus this is found to be a minimum 
when D 2 = \ . B 2 , i.e. when M is in the W.L. or where it crosses 
the 45 line. 

The M curve is a hyperbola referred to the vertical at zero 
draught, and the C.B. line as axes, having the equation 



the axes being asymptotes. 

2. A vessel with a triangular section, vertex down. This is 
dealt with on p. 117, where it is seen that the M curve is 
straight 

BM = J.^. M from base = f .D + J.jj 

M from base B 2 

/. - g - f + f-- jJJ I + I tan a = constant 

Tt 

Pj = tan a, where a is the semi-vertical angle 

i.e. M curve is a straight line, making an angle 0, with the 
base line such that tan = (i + tan 2 a). 

3. Vessel with parabolic section. A parabola has the equa- 
tion referred to axes at the vertex, y 2 = 4ax, i.e. for x draught 
breadth at waterline is 2y = 4* ax (Fig. 6oA). 

The C.G. of a parabola is f the depth, so that the C.B. 
locus is a straight line making an angle with the base of 
tan^d). Area of parabola = f . circumscribing rectangle. 



BM = f.2.lf = * ' D = = C nStant 

i.e. the locus of metacentres in metacentric diagram is straight 
and parallel to the C.B. locus. 



Conditions of Equilibrium, Transverse Metacentre, etc. 133 

4. Vessel with circular section (Fig. 6oB). In this case the 
metacentre is always at the centre, so that the M curve is a 
straight line at mid depth. 




FIG. 6oA. 



FIG. 6oB. 



The B curve is a flat curve starting at an angle with the 
base, such that = tan" 1 ^), since there the circle may be 
regarded as a parabola. 

For mid depth B below W.L. is -- (a being radius), and 

3^ 

the inclination of tangent is an angle a such that tan a = '54 
by the formula given on p. 118. 

When completely immersed the curve finishes as a tangent 
to the M curve. 

Curves of Buoyancy, etc. The surface of biioyancy for a 
given displacement is the surface traced out by the centre of 
buoyancy as the vessel takes up all possible positions while 
maintaining that displacement. 

The surface of flotation is the surface traced out by the centre 
of flotation under the same conditions. 

The curve of buoyancy is the curve traced out on the transverse 
vertical plane by the projection of the centre of buoyancy as the 
ship is continually revolved about a longitudinal axis fixed in 
direction while maintaining the same displacement. This curve is 
also termed an isovol. 

The curve of flotation is the curve traced out by the projection 
of the centre of flotation under the same conditions. 



134 Theoretical Naval Architecture. 

The pro-metacentre is the intersection of any two consecutive 
lines of action of buoyancy, as M' in Fig. 6oE. When consecutive 
lines do not intersect the pro-metacentre is the intersection of one 
of them with the common perpendicular. For a condition of 
equilibrium this intersection of consecutive lines of buoyancy is 
the metacentre. 

The metacentric is the locus of pro-metacentres. 

The following are definitions of various sorts of equilibrium: 

(1) Rotation in a given direction only. 

(a) Stable equilibrium for a given direction of inclination 
when, on being slightly displaced in that direction 
from its position of rest, the vessel tends, on being 
released, to go back to that position. 

() Unstable equilibrium is as (a), only that the vessel moves 
further from the position of rest. 

(c) Indifferent or unstable equilibrium the vessel neither 
tends to return to or to go further from the position of 
rest. 

(rf) Mixed Equilibrium if stable for one direction of in- 
clination and unstable for the opposite direction. 

(2) Rotation in all directions. 

(a) Absolute equilibrium when only stable or unstable for 
any direction of inclination. 

(b) Relative stability when stable in some directions and 
unstable in others. 

Thus, in a ship 

(i) If the C.G. is below the transverse metacentre M T , she is 

absolutely stable, 
(ii) If the C.G. is above the longitudinal metacentre M L , 

she is absolutely unstable. 

(iii) If the C.G. is between M T and M L , she has relative 
stability, being stable for longitudinal inclinations and 
unstable for transverse inclinations. 
The above definitions are well illustrated by a floating cube of 

s.g. i. 

(a) When floating with a face horizontal, the cube is 
absolutely unstable. 

(b] With one corner downwards, the cube has absolute 
stability. 

In going from one point on the surface of buoyancy to the con- 
secutive point, B to B', BB' = ^ y^% and BB' is parallel to g^. 
Hence, in the limit BB' is parallel to the water-plane, so that the 



Conditions of Equilibrium, Transverse Metacentre, etc. 135 



tangent plane at any point to the surface of buoyancy is parallel to 
the corresponding water-plane, and the normal to it through the 
point of contact gives the line of action of buoyancy. The surface 
must be wholly convex to the tangent plane or wholly concave to 
some interior point. Similar reasoning will also apply to the curve 
of buoyancy. 

For a position of equilibrium, the line of action of the buoyancy 
must pass through the C.G. ; therefore, the number of positions of 
equilibrium that a body can take up is equal to the number of 
normals that can be drawn from the C.G. to the surface of buoyancy. 
For a given direction of inclination the number of positions of equi- 
librium equals the number of normals that can be drawn from the 
C. G. to the curve of buoyancy. 

When BG thus drawn is a minimum, the equilibrium is stable. 

When BG thus drawn is a maximum, the equilibrium is 

unstable ; 

for the stability is the same as that of the curve of buoyancy 
rolled along a smooth horizontal plane, the weight being concen- 
trated at the C.G. In moving from one position of equilibrium to 
another, if the C.G. has to be raised we have stable equilibrium, 
i.e. B'G > BG. If unstable, similarly B'G < BG. 

The centre of curvature of the surface of buoyancy is what we 
have termed the pro-metacentre, and the radius of curvature is 

given by R =^ where I is the moment of inertia of the water- 
plane about an axis through its C.G. perpendicular to the plane of 
rotation, and V is the volume of 
displacement. This is proved 
exactly as in Chap. III. for the 
upright BM. 

Lederfs theorem for the radius 
of curvature of the curve of 
flotation. 

In Fig. 6oc WL and W'L' are 
consecutive water-lines for the 
upright ivl, iv'l', when inclined to 
a small angle, the increment of 
displacement being AV. Then 
when inclined the buoyancy V 

acts through M, and that of AV through O, the centre of curva- 
ture of the curve of flotation. B and B' are the upright C.B.'s 
and M' the metacentre for the water-line W'L', and V + AV acts 
through M' for a small inclination. 




i 
FIG. 6oc. 



136 Theoretical Naval Architecture. 

Taking moments about C, the C.G. of the layer AV, we 
have 

(V x CM) + (AV x CO) = (V + AV)CM' 
or V(BM - BC) + (AV x CO) = (V + AV)(B'M' - B'C) 

now V x BC = (V + AV)B'C 

so that (V x BM) + (AV x CO) = (V + AV)B'M' 
or I + (AV x CO) = I + Al 

d\ 
i.e. OC = -jr.. in the Limit, which is 

the expression for the radius of curvature of the curve of flotation 
usually called r. 

It can be readily shown that if a weight be added at the point 
O the moment of initial stability is not changed. For ordinary 
ships parallel-sided at the water-line d\ is zero or practically so, so 
that O is in the water-line. We may therefore say that, generally 
speaking, if a weight is added above the water-line it will diminish 
the stability ; if added at the water-line there is no change in 
the stability ; if added below the water-line the stability is 
increased. 

Examples. (i.) r for a body of rectangular section, r o. 

(ii.) r for a body of triangular equilateral section, angle 20. 

(a) corner downwards, r d tan 2 (d = draught). 

(b) corner upwards, r = c tan 2 (c being dis- 

tance of water-line from vertex). 

(iii.) r for a circular section, radius a, r = a cos 6. 
(26 being angle subtended at the centre by the water-line.) 

(iv.) Show that an added weight to keep the metacentric 
height constant should be placed the same distance 
from G as O is distant from M. 

(v.) In Example (ii.) () above, if tan 6 = f and depth is 40' 
then if the draught is less than 14*4 ft. a small addition 
of ballast to the base of the triangle will make the 
body less stable, but at greater draughts the stability 
increases with the addition of ballast. 



GEOMETRY OF THE METACENTRIC DIAGRAM. 

r. Tangent to the curve of C.B. In Fig. 6oD, let be the 
inclination of this locus to the horizontal at water-line WL. 

Then for an increment of displacement AV and of draught by 

the C.B. will rise an amount ^ . AV = ' . Ay. 



Conditions of Equilibrium, Transverse Metacentre, etc. 1 37 



tan e = IL s - e -2O? = A- h W here A = area of water-plane, 

h = C.B. below water-line, 
V = volume displacement. 



w 



w 



V+AV 



Mr 




FIG. 600. 

Examples. (i.) for a box-shaped vessel, tan 6 = O'5- 
(ii.) for a triangular section, tan = O'66. 
(iii.) for a circular section at) , Q 

half depth } tan = o 54. 

For ordinary ships it is found that tan Q = 0*55 about. 

2. Tangent to the curve of metacentres. The increment of 
volume AV, for a small inclination, has its line of action through O, 
the centre of curvature of the curve of flotation. Let OM = k 
then (V + AV)MM' = AV x OM . = AV x k (Fig. 6oc) 



MM' = 



V 



Ay. 



If $ be the angle the tangent to the M curve makes with the 
horizontal, then 

rise or fall of M A . k . 
tan <}> = - - = - 



in the limit 



If k = o, then tan </> = o and the M curve is horizontal, i.e. when 
the M curve is horizontal, the points M and O coincide. This is 
otherwise obvious, as the added buoyancy will act through M, which 
is therefore fixed in height for a small increment of draught. In a 
box-shaped ship the M curve is horizontal when D 2 = \ B 2 . 

Co-ordinates of the Centre of Buoyancy referred to axes through 
the itpright C.B. 

In Fig. 6oE, x and y are the co-ordinates of B, the C.B. at 
angle 0. For an increment of angle dQ, B' is the new C.B. and 



133 



Theoretical Naval Architecture. 



x + dx, y 4- tf^the co-ordinates, BM', B'M' the normals at B and B' 
intersect in M' the pro-metacentre, and BM' = B'M' = R, and 

R = ^ . BB' = R . dQ, and dx - BB' . cos 6, dy = BB' . sin 0, 

.*. dx = R . cos e . d9 dy = R . sin . dQ 

and x = JR . cos 6 . d9 y - /R . sin e . d9. 

Curves of R cos 6, R . sin can be drawn on a 6 base and integrated 
up to the various angles. Thus, x and_x can be obtained and so 
the curve of buoyancy drawn in. The righting arm at angle is 
given by GZ = .r.cos 6 + y sin 9 B G.sin e. 





FIG. 6oE. 



FIG. 6oF. 



This is the French method of calculating stability due to 
M. Reech 

For a box so long as wall sided R = B M . sec 2 6 
so that x JB M . sec 3 & cos 6 dQ B M . tan 9 

y = JB M sec 3 B . sin 6 dQ 

= j*B M . sec 2 6 tan d6 - B M . tan 2 

This is the solution of question 35 in the Appendix. 
Question 36 is solved as follows : 

G' the new C.G. will lie on BM' and GG' = ^^ 

GG' cos 6 = x cos 6 + y sin B B G sin 6 
GG' x +y . tan B G . tan 6 

= B M . tan + JB M . tan 3 e - B G . tan 6 
= GM . tan e + *B M . tan 3 6 



Also 



GG' = 



w X d 
W 



f 



Conditions of Equilibrium, Transverse Metacentre, etc. 1 39 

STABILITY OF A WALL-SIDED VESSEL, 

, v x hh 1 

In Fig. 6oF, BR = = 

v ~ \ -y 2 - tan e 
hh' f the projection of aa on to W'L' 

= ^ the projection of SL 4- La on to W'L' 
= ^ (_y . cos + ^ .y . tan . sin 0) 



:= = BM . sin 0(i + tan 2 0) taking a unit length 



= BR also. 
GZ = BR - EG. sin 

= BM sin + BM . tan 2 sin - BG sin 
= sin 0(GM + |BM . tan 2 0). 

This can be used to construct the curve of stability (see Chap. 
V.) so far as the ship is wall-sided above and below water, and 
can be used to check the cross curve at 15 say obtained by the 
Integrator. 

This formula may be used to determine the angle to which a 
ship with negative metacentric height will loll over, for GZ will 
then be zero, and we have 



and the metacentric height when at the angle is 2 -. (See 

example 37 in Appendix.) 

EXAMPLES TO CHAPTER III. 

1. Find the circular measure of 5 , ioj, 15!. 

Ans. 0-09599; 0-17889; 0-27489. 

2. Show that sin 10 is one-half per cent, less in value than the circular 
measure of 10, and that tan 10 is one per cent, greater in value than the 
circular measure of 10. 

3. A cylinder weighing 500 Ibs., whose centre of gravity is 2 feet from 
the axis, is placed on a smooth table and takes up a position of stable 
equilibrium. It is rolled along parallel to itself through an angle of 60. 
What will be the tendency then to return to the original position ? 

Ans. 866 foot-lbs. 

4. Find the moment of inertia about the longest axis through the centre 
of gravity of a figure formed of a square of side 20, having a semicircle at 
each end. 



140 Theoretical Naval Architecture. 

5. Find the moment of inertia of a square of side 2a about a diagonal. 

Am. \ a\ 

6. A square has a similar square cut out of its centre such that' the 
moment of inertia (about a line through the centre parallel to one side) of 
the small square and of the portion remaining is the same. What pro- 
portion of the area of the original square is cut out ? 

Ans. 071 nearly. 

7. A vessel of rectangular cross-section throughout floats at a constant 
draught of 10 feet, and has its centre of gravity in the load water -plane. 
The successive half-ordinates of the load water-plane in feet are o'5, 6, 12, 
16, 15, 9, o; and the common interval 20 feet. Find the transverse 
metacentric height. 

Ans. 8 inches. 

8. A log of fir, specific gravity 0*5, is 12 feet long, and the section is 
2 feet square. What is its transverse metacentric height when floating in 
stable equilibrium in fresh water ? 

Ans. o - 47 foot. 

9. The semi-ordinates of a water-plane 34 feet apart are 0*4, 13*7, 
25*4, 32-1, 34-6, 35-0, 34-9, 34-2, 32-1, 23-9, 6-9 feet respectively. Find 
i-ts moment of inertia about the centre line. 

Ans. 6,012,862. 

10. The semi-ordinates of the load water-plane of a vessel are o, 3*35, 
6-41, 8-63, 9-93, 10-44, I0'37, 9'94 8-96, 7-16, and 2-5 feet respectively. 
These ordinates being 21 feet apart, find 

(1) The tons per inch immersion. 

(2) The distance between the centre of buoyancy and the transverse 

metacentre, the load displacement being 484 tons. 

Ans. (i) 773 tons; (2) 5*2 feet nearly. 

11. The semi-ordinates, l6'6 feet apart, of a vessel's water-plane are 
0-2, 2-3, 6-4, 9-9, 12-3, 13-5, 13-8, 137, 12-8, 10-6, 6-4, i'9, 0-2 feet 
respectively, and the displacement up to this water-plane is 220 tons. Find 
the length of the transverse BM. 

Ans. 2O'6 feet. 

12. A vessel of 613 tons displacement was inclined by moving 30 cwt. 
of rivets across the deck through a distance of 22' 6". The end of a plumb- 
line 10 feet long moved through 2\ inches. What was the metacentric 
height at the time of the experiment ? 

Ans. 2 -93 feet. 

13. The semi-ordinates of a ship's water-plane 35 feet apart are, com- 
mencing from forward, 0*4, 7*12, 15*28, 2i'8, 25-62, 26-9, 26-32, 24-42, 
20 '8, I5'I5? 6 -39 feet respectively. There is an after appendage of 116 
square feet, with its centre of gravity 180 feet abaft the midship ordinate. 
Find 

(1) The area of the water-plane. 

(2) The tons per inch immersion. 

(3) The distance of the centre of flotation abaft amidships. 

(4) The position of the transverse metacentre above the L.W.L., taking 

the displacement up to the above line as 5372 tons, and the 
centre of buoyancy of this displacement 8'6i feet below the 
L.W.L. 

Ans. (I) 1 3, 292 square feet; (2) 31-6 tons; (3) 14*65 feet ; (4) 3-34 
feet. 

14. A ship displacing 9972 tons is inclined by moving 40 tons 54 feet 



Conditions of Equilibrium, Transverse Metacentre, etc. 1 4 1 

across the deck, and a mean deviation of 9^ inches is obtained by pendulums 
15 feet long. Find the metacentric height at the time of the operation. 

Ans. 4* 1 8 feet. 

15. A ship weighing 10,333 tons was inclined by shifting 40 tons 52 
feet across the deck. The tangent of the angle of inclination caused was 
found to be 0*05. If the transverse metacentre was 475 feet above the 
designed L. W.L., what was the position of the centre of gravity of the ship 
at the time of the experiment ? 

Ans. 073 foot above the L.W.L. 

1 6. A vessel of 26 feet draught has the moment of inertia of the L. W. P. 
about a longitudinal axis through its centre of gravity 6, 500,000 in foot- 
units. The area of the L.W.P. is 20,000 square feet, the volume of dis- 
placement 400,000 cubic feet, and the centre of gravity of the ship may be 
taken in the L.W.P. Approximate to the metacentric height. 

Ans. 5^ feet. 

17. Prove the rule given on p. 62 for the distance of the centre of 

gravity of a semicircle of radius a from the diameter, viz. ^-a, by finding 

the transverse BM of a pontoon of circular section floating with its axis in 
the surface of the water. 

(M in this case is in the centre of section.) 

1 8. Take a body shaped as in Kirk's analysis, p. 84, of length 140 
feet ; length of parallel middle body, loo feet'; extreme breadth, 30 feet ; 
draught, 12 feet. Find the transverse BM. 

Ans. 57 feet. 

19. A vessel of 1792 tons displacement is inclined by shifting 5 tons 
already on board transversely across the deck through 20 feet. The end 
of a plumb-line 15 feet long moves through 5J inches. Determine the 
metacentric height at the time of the experiment. 

Ans. 1-91 feet. 

20. A vessel of displacement 1722 tons is inclined by shifting 6 tons of 
ballast across the deck through 22 \ feet. A mean deviation of loj inches 
is obtained with pendulums 15 feet long. The transverse metacentre is 
15*28 feet above the keel. Find the position of the centre of gravity of the 
ship with reference to the keel. 

Ans. 13*95 f et 

21. The ship in the previous question has 169 tons to go on board at 
lo feet above keel, and 32 tons to come out at 20 feet above keel. Find 
the metacentric height when completed, the transverse metacentre at the 
displacement of 1859 tons being 15-3 feet above keel. 

Ans. i -8 feet. 

22. A vessel of 7000 tons displacement has a weight of 30 tons moved 
transversely across the deck through a distance of 50 feet, and a plumb-bob 
hung down a hatchway shows a deviation of 12 inches in 15 feet. What 
was the metacentric height at the time of the operation ? 

Ans. 3 '2 1 feet. 

23. A box is 200 feet long, 30 feet broad, and weighs 2000 tons. Find 
the height of the transverse metacentre above the bottom when the box is 
floating in salt water on an even keel. Ans. 12 '26 feet. 

24. Show that for a rectangular box floating at a uniform draught of d 
feet, the breadth being 12 feet, the distance of the transverse metacentre 

above the bottom is given by 24 feet, and thus the transverse meta- 
centre is in the water-line when the draught is 4-9 feet. 



142 Theoretical Naval Architecture. 

25. A floating body has a constant triangular section. If the breadth 
at the water-line is A/ 2 times the draught, show that the curve of metacentres 
in the metacentric diagram lies along the line drawn from zero draught at 
45 to the horizontal, and therefore the metacentre is in the water-line for 
all draughts. 

26. A floating body has a square section with one side horizontal. 
Show that the transverse metacentre lies above the centre of the square 
so long as the draught does not much exceed 21 per cent, of the depth of 
the square. Also show that as the draught gets beyond 21 per cent, of the 
depth, the metacentre falls below the centre and remains below until 
the draught reaches 79 per cent, of the depth ; it then rises again above 
the centre of the square, and continues to rise as long as any part of the 
square is out of the water. 

(This may be done by constructing a metacentric diagram, or by using the 
methods of algebra, in which case a quadratic equation has to be solved.) 

27. Show that a square log of timber of 12 inches side, 10 feet long, and 
weighing 320 Ibs., must be loaded so that its centre of gravity is more than 
I inch below the centre in order that it may float with a side horizontal 
in water of which 35 cubic feet weigh I ton. 

28. A prismatic vessel is 70 feet long. The section is formed at the 
lower part by an isosceles triangle, vertex downwards, the base being 20 
feet, and the height 5 feet ; above this is a rectangle 20 feet wide and 5 feet 
high. Construct to scale the metacentric diagram for all drafts. 

29. A vessel's load water-plane is 380 feet long, and 75 feet broad, and 
its moment of inertia in foot-units about the centre line works out to 
8,000,000 about. State whether you consider this a reasonable result to 
obtain, the water-plane not being very fine. 

TJJ 

30. Find the value of the coefficient a in the formula BM = a 

referred to on p. in, for floating bodies having the following sections 
throughout their length : 

(a) Rectangular cross-section. 
(6) Triangular cross-section, vertex down. 

(c) Vertical-sided for one half the draught, the lower half of the section 
being in the form of a triangle. 

Arts, (a) 0-08; (6) c-i6 ; (c) o'li. 
For ordinary ships the value of a will lie between the first and last of these. 

31. A lighter in the form of a box is 100 feet long, 20 feet broad, and 
floats at a constant draught of 4 feet. The metacentric height when empty 
is 6 feet. Two bulkheads are built 10 feet from either end. Show that a 
small quantity of water introduced into the central compartment will render 
the lighter unstable in the upright condition. 

32. At one time, in ships which were found to possess insufficient sta- 
bility, girdling was secured to the ship in the neighbourhood of the water- 
line. Indicate how far the stability would be influenced by this means. 

33. A floating body has a constant triangular section. If the breadth 
at the water-line is equal to the draught, show that the locus of metacentres 
in the metacentric diagram makes an angle with the horizontal of about 40. 

34. A cylinder is placed into water with its axis vertical. Show that if 
the centre of gravity is in the water-plane, the cylinder will float upright if 
the radius -r- the draught is greater than */2. 

35. In a wholly submerged body show that for stable equilibrium the 
centre of gravity must lie below the centre of buoyancy. 

36. A floating body has a constant triangular section, vertex down- 
wards, and has a constant draught of 12 feet, the breadth at the water-line 



Conditions of Equilibrium, Transverse Metacentre, etc. 143 



being 24 feet. The keel just touches a quantity of mud, specific gravity 2. 
The water-level now falls 6 feet : find the amount by which the meta- 
centric height is diminished due to this. 1 Ans. 2f feet about. 

37. A floating body of circular section 6 feet in diameter has a meta- 
centric height of I "27 feet. Show that the centre of buoyancy and centre 
of gravity coincide, when the body is floating with the axis in the surface. 

38. It is desired to increase the metacentric height of a vessel which is 
being taken in hand for a complete overhaul. Discuss the three following 
methods of doing this, assuming the ship has a metacentric diagram as in 
Fig. 56, the extreme load draught being 15 feet : 

(1) Placing ballast in the bottom. 

(2) Removing top weight. 

(3) Placing a girdling round the ship in the neighbourhood of the 

water-line. 

39. Show that the angle in Fig. 56 is between 29 and 30 for a 
vessel whose coefficient of L.W.P. is 075, and whose block coefficient 
of displacement is 0-55. In any case, if these coefficients are denoted by 

n and k respectively, show that tan = \ + ^ approximately (use Mor- 
rish's formula, p. 65). 

40. From the following information construct the metacentric diagram, 
using a scale of inch = I foot, and state the metacentric height and 
draught in the three conditions given. 



Draught. 


Displacement 
in tons. 


Tons per 
inch. 


C.B. below 
i9-foot WL. 


BM. 


2i ; 9 ;: 

19' o" 


5256 
4383 


27-I 
26-48 


6-25' 

7-8' 


8-85' 
10-4' 


16' 3" 


3527 


25'37 


9'35' 


12-2' 


13' 6" 


2714 


23-84 


10-9' 


14-5' 



(1) Deep load 1000 tons coal 5030 tons, C.G. o'3 feet below 19' WL. 

(2) Normal load 400 4430 0-35 

(3) Light condition 3915 ,, 0*3 above ,, 

Ans. (i) 2-9', 21' of'; (2) 2-95', 19' if" ; (3) 2' 4 ', 1/6" 

41. A weight of 10 tons is shifted 40 ft. transversely across the deck 
of a vessel having a compartment partially filled with salt water, the free 
surface of this water being 25 ft. long by 50 ft. uniform breadth. Calculate 
the heel in degrees, having given displacement of vessel 8000 tons, C.G. of 
ship and contained water 15 ft. above keel. Transverse M. i6 ft. above 
keel. (Durham B.Sc. 1910). 

The actual GM is i feet, but the virtual GM is less than this by the 

amount ^r, where * is the moment of inertia of free water surface about 
a longitudinal axis through its C.G. 

* = & 2 S.So- 50* V = 8000 X 35 

so that ^ = 0-93 ft. The virtual GM is therefore 0-57 ft., 9 being angle 
of heel. 



W x GM X sin = w X d 



or sin = 



10 X 40 



= 0-088 



8000 X 0-57 

and = 5 degrees. 



1 This example is worked out at the end of the Appendix. 



CHAPTER IV. 

LONGITUDINAL METACENTRE, LONGITUDINAL BM, 
CHANGE OF TRIM. 

Longitudinal Metacentre. We now have to deal with 
inclinations in a fore-and-aft or longitudinal direction. We 
do not have the same difficulty in fixing on the fore-and-aft 
position of the centre of gravity of a ship as we have in fixing 
its vertical position, because we know that if a ship is floating 
steadily at a given water-line, the centre of gravity must be in 
the same vertical line as the centre of buoyancy, by the con- 
ditions of equilibrium laid down on p. 93. It is simply a 
matter of calculation to find the longitudinal position of the 
centre of buoyancy of a ship when floating at a certain water- 
line, if we have the form of the ship given, and thus the fore- 
and-aft position of the centre of gravity is determined. 

We have already dealt with the inclination of a ship in a 
transverse direction, caused by shifting weights already on 
board across the deck ; and in a precisely similar manner we 
can incline a ship in a longitudinal or fore-and-aft direction by 
shifting weights along the deck in the line of the keel. The 
trim of a ship is the difference between the draughts of water 
forward and aft. Thus a ship designed to float at a draught 
forward of 12 feet, and a draft aft of 15 feet, is said to trim 3 feet 
by the stern. 

We have, on p. 97, considered the definition of the trans- 
verse metacentre, and the definition of the longitudinal meta- 
centre is precisely analogous. 

For a given water-line WL of a vessel, let B be the centre 
of buoyancy (see Fig. 61), and BM the vertical through it. 



Longitudinal Metacentre, Longitudinal BM, etc. 145 

Suppose the trim of the vessel to change slightly, 1 the vessel 
retaining the same volume of displacement, B' being the new 
centre of buoyancy, and B'M the vertical through it, meeting 




FIG. 61. 

BM in M. Then the point M is termed the longitudinal 
metacentre. 

The distance between G, the centre of gravity of the ship, 
and M, the longitudinal metacentre, is termed the longitudinal 
metacentric height. 

Formula for finding the Distance of the Longi- 
tudinal Metacentre above the Centre of Buoyancy. 
Let Fig. 62 represent the profile of a ship floating at the water- 
line WL', the original water-line being WL. The original 
trim was AW - BL ; the new trim is AW - BL'. The change 
of trim is 

(AW - BL) - (AW - BL') = WW + LL' 

i.e. the change of trim is the sum of the changes of draughts 
forward and aft. This change, we may suppose, has been 
caused by the shifting of weights from aft to forward. The 
inclination being regarded as small, and the displacement 
remaining constant, the line of intersection of the water-planes 
WL, W'L' must pass through the centre of gravity of the water- 
plane WL, or, as we have termed it, the centre of flotation, 
in accordance with the principle laid down on p. 98. This 
centre of flotation will usually be abaft the middle of length, 
and this introduces a complication which makes the calculation 
for the longitudinal metacentre more difficult than the corre- 
1 Much exaggerated in the figure. 



146 



Theoretical Naval Architecture. 



spending calculation for the transverse metacentre. In this 
latter case, it will be remembered that the centre of flotation is 
in the middle line of the water-plane. 




FIG. '62. 

In Fig. 62 

Let B be the centre of buoyancy when floating at the 

water-line WL ; 

B', the centre of buoyancy when floating at the water- 
line WL' ; 

FF, the intersection of the water-planes WL, WL' ; 
v, the volume of either the immersed wedge FLL' or 

the emerged wedge FWW ; 
-, ^, the centres of gravity of the wedges WFW, LFL 

respectively ; 

V, the volume of displacement in cubic feet ; 
0, the angle between the water-lines WL, WL', which 
is the same as the angle between BM and B'M 
(this angle is supposed very small). 
We have, using the principle laid down on p. 100 
v X = V v BB' 



Longitudinal Metacentre, Longitudinal BM, etc. 147 

, r PR' 
or BB = 

But BB' = BM x 6 (0 is in circular measure) 



The part of this expression that we do not know is v X gg\ 
or the moment of transference of the wedges. At P take a 
small transverse slice of the wedge FLL', of breadth in a fore- 
and-aft direction, dx\ length across, 2y; and distance from 
F, x. Then the depth of the slice is 

x x 

and the volume is 2y x xO X dx 

This is an elementary volume, analogous to the elementary 
area y . dx used in finding a large area. The moment of this 
elementary volume about the transverse line FF is 

2yx . . dx X x 
or 2yx* .6 .dx 

If we summed all such moments as this for the length FL, 
we should get the moment v X F^', and for the length FW, 
v X F-, or for the whole length, v X gg' j therefore, using our 
ordinary notation 

X gg = f2yx* .O.dx 

= 26jyx*.dx (6 being constant) 

We therefore have 



or 



Referring to p. 103, it will be seen that we denned the 
moment of inertia of an area about a given axis as 

JWA X/ 

where dA. is a small elementary area ; 

y its distance from the given axis. 

Consider, now, the expression obtained, 2fyx? . dx. The 
elementary area is 2y . dx, and x is its distance from a 



148 



Theoretical Naval Architecture. 



transverse axis passing through the centre of flotation. We 
may therefore say 



where I is the moment of inertia of the water-plane about a 
transverse axis passing through the centre of flotation. It will 
be seen at once that this is the same form of expression as for 
the transverse BM. 

The method usually adopted for finding the moment of 
inertia of a water-plane about a transverse axis through the 
centre of flotation is as follows * : 

We first find the moment of inertia about the ordinary 
midship ordinate. If we call this I, and y the distance of the 
centre of flotation from the midship ordinate, we have, using 
the principle given on p. 104 

I = I + A/ 
or I = I - A/ 

The method actually adopted in practice will be best under- 
stood by working the following example. 



Numbers 
of 
ordinates. 


Semi- 
ordinates 
ofL.W.P. 


Simpson's 
multi- 
pliers. 


Products 
for area. 


Multi- 
pliers for 
moment. 


Products 
for 
moment.^ 


Multi- 
pliers for 
moment 
of inertia. 


Products 
for 
moment 
of inertia. 


I 


O'O 


* 


O'O 


5 


o-o 


5 


O'O 


I* 


i'37 


2 


274 


4i 


12-33 


4i 


5 5 '49 


2 


2-67 


ii 


4-01 


4 


16-04 


4 


64-16 


3 


4-87 


4 


19-48 


3 


58-44 


3 


i75'32 


4 


6-31 


2 


12-62 


2 


25-24 


2 


50-48 


5 


6-85 


4 


27-40 


I 


27-40 


I 


27-40 


6 


7-21 


2 


14-42 


o 


I39-45 


O 





7 


7*15 


4 


28-60 


I 


28-60 


I 


28-60 


g 


6-87 


2 


1374 


2 


27-48 


2 


54-96 


9 


6'33 


4 


25-32 


3 


75^6 


3 


227-88 


10 


5-08 


I* 


7-62 


4 


30-48 


4 


121-92 


10* 


3-56 


2 


7-12 


4* 


32-04 


4 


144-18 


ii 


071 


i 


o'35 


5 


175 


5 


8-75 


163-42 196-31 9S9'i4 


= Si I 39'45 = S, 



56-86 =S 2 



1 This calculation for the L.W.P. is usually performed on the displace- 
ment sheet. 



Longitudinal Metacentre, Longitudinal BM y etc. 149 

In column 2 of the table are given the lengths of semi- 
ordinates of a load water-plane corresponding to the numbers 
of the ordinates in column i. The ordinates are 7*1 feet 
apart. It is required to find the longitudinal BM, the dis- 
placement being 91-6 tons in salt water. 

The distance apart of the ordinates being 7*1 feet, we have 

Area = 163*42 X (J X 7'i) X 2 

= 7 7 3' 5 square feet 

Distance of centre of gravity of 1 __ 56*86 X 7'i 
water-plane abaft No. 6 ordinate J " 163*42 2 4 ^ 

(the stations are numbered from forward). 

The calculation up to now has been the ordinary one 
for finding the area and position of the centre of gravity. 
Column 4 is the calculation indicated by the formula 

Area = 2Jy . dx 

Column 6 is the calculation indicated by the formula 
Moment = 2Jyx . dx 

It will be remembered that in column 5 we do not put 
down the actual distances of the ordinates from No. 6 ordinate, 
but the number of intervals away; the distance apart of the 
ordinates being introduced at the end. By this means the 
result is obtained with much less labour than if column 5 
contained the actual distances. The formula we have for the 
moment of inertia is 2Jy . x* . dx. We follow a similar process 
to that indicated above ; we do not multiply the ordinates by 
the square of the actual distances, but by the square of the 
number of intervals away, leaving to the end the multiplication 
by the square of the interval. Thus for ordinate No. 2 the 
actual distance from No. 6 is 4X7*1 = 28 "4 feet. The 
square of this is (4) 2 X (7*i) 2 . For ordinate No. 4 the square of 
the distance is (2) 2 X (7*i) 2 . The multiplication by (7*i) 2 can 
be done at the end. In column 7 is placed the number of 
intervals from No. 6, as in column 5 ; and if the products in 
column 6 are multiplied successively by the numbers in 
column 7, we shall obtain in column 8 the ordinates put 



15 Ttteoretical Naval Architecture. 






through Simpson's rule, and also multiplied by the square of 
the number of intervals from No. 6 ordinate. The whole of 
column 8 is added up, giving a result 959*14. To obtain the 
moment of inertia about No. 6 .ordinate, this has to be multi- 
plied as follows : 

(a) By one-third the common interval to complete Simp- 

son's rule, or \ X 7*1. 

(b) By the square of the common interval, for the reasons 

fully explained above. 

(c) By two for both sides. 

We therefore have the moment of inertia of the water-plane 
about No. 6 ordinate 

959-14 x ( X 7-1) X (7'i) 2 X 2 = 228,858 
The moment of inertia about a transverse axis through the 
centre of flotation will be less than this by considering the 
formula I = I + Ay, where I is the value found above about 
No. 6 ordinate, and I is the moment of inertia we want. We 
found above that the area A = 773*5 square feet, and y = 2*47 
feet; 

.Mo =228,858 -(773*5 X 2- 47 2 ) 
= 224,1391 

The displacement up to this water-plane is 91*6 tons, and 
the volume of displacement is 

91*6 x 35 = 3206 cubic feet 
The longitudinal BM = 



3206 

Approximate Formula for the Height of the Longi- 
tudinal Metacentre above the Centre of Buoyancy. 
The following formula is due to M. J. A. Normand, M.I.N.A., 2 
and is found to give exceedingly good results in practice : 

Let L be the length on the load water-line in feet ; 
B, the breadth amidships in feet ; 

1 See note at end of chapter, p. 167. 

8 See "Transactions of the Institution of Naval Architects, ' 1882. 



Longitudinal Metacentre, Longitudinal BM, etc. 151 

V, the volume of displacement in cubic feet ; 
A, the area of the load water-plane in square feet. 
Then the height of the longitudinal metacentre above the centre 
of buoyancy 

A 2 XL 
H = 0-0735B1TV 

In the example worked above, the breadth amidships was 
14*42 feet; and using the formula, we find 

H = 6 7 '5 feet nearly 

This compares favourably with the actual result of 69*9 feet. 
The quantities required for the use of the formula would all be 
known at a very early stage of a design and a close approxima- 
tion to the height H can thus very readily be obtained. A 
formula such as this is useful as a check on the result of the 
calculation for the longitudinal BM. 

We may also obtain an approximate formula in the same 
manner as was done for the transverse BM on p. in. Using 
a similar system of notation, we may say 

Moment of inertia of L.W.P. about a trans- 



T * v 

verse axis through the centre of flotation f 

n f being a coefficient of a similar nature to n used on p. 107. 

Volume of displacement = xLxBxD 

tt X L 3 X B 
"' ~ 



where b is a coefficient obtained from the coefficients n' and k. 
Sir William White, in the " Manual of Naval Architecture," says, 
with reference to the value of , that "the value 0-075 ma y be 
used as a rough approximation in most cases ; but there are 
many exceptions to its use." If this approximation be applied 
to the example we have worked, the mean moulded draught 
being 5*8 feet 

The value of H = 65 feet 



152 Theoretical Naval Architecture. 

This formula shows very clearly that the length of a ship is 
more effective than the draught in determining the value of the 
longitudinal BM in any given case. For vessels which have an 
unusual proportion of length to draught, the values of the longi- 
tudinal BM found by using this formula will not be trustworthy. 

To estimate the Displacement of a Vessel when 
floating out of the Designed Trim. The following 
method is found useful when it is not desired to actually 
calculate the displacement from the drawings, and a close 
approximation is sufficiently accurate. Take a ship floating 
parallel to her designed L.W.L. ; we can at once determine 
the displacement when floating at such a water-line from the 
curve of displacement (see p. 25). If now a weight already 
on board is shifted aft, say, the ship will change trim, and she 
will trim more by the stern than designed. The new water- 
plane must pass through the centre of gravity of the original 
water-plane, or, as we have termed it, the centre of flotation, and 




FIG. 63. 

the displacement at this new water-line will be the same as at 
the original water-line. Now, when taking the draught of water 
a vessel is actually floating at, we take the figures set up at or 
near the forward and after perpendiculars. These draughts, 
if not set up at the perpendiculars, can be transferred to the 
perpendiculars by a simple calculation. The draughts thus 
obtained are added together and divided by two, giving us 
the mean draught. Now run a line parallel to the designed 
water-line at this mean draught, as in Fig. 63, where WL 
represents the actual water-line, and wl the line just drawn. 
It will not be true that the displacement of the ship is the same 
as that given by the water-line wl. Let F be the centre of 
flotation of the water-line wt, and draw WL' through F parallel 
to WL. Then the actual displacement will be that up to WL', 
which is nearly the same as that up to wl, with the displacement 



Longitudinal Metacentre, Longitudinal J3M, etc. 153 

of the layer WW'L'L added. The displacement up to wl is 
found at once from the curve of displacement. Let T be the 
tons per inch at //, and therefore very nearly the tons per inch 
at W'L' and WL. SF, the distance the centre of flotation of 
the water-plane wl is abaft the middle of length, is supposed 
known, and equals d inches, say. Now, the angle between wl 
and WL is given by 

tan0 = 



length of ship 
_ amount out of normal trim 
length of ship 

But if x is the thickness of layer in inches between W'L' and 
WL, we also have in the triangle SFH 

tan 6 = -j very nearly (for small angles tan = sin 
very nearly) 

and accordingly x may be determined. This, multiplied by 
the tons per inch T, will give the displacement of the layer. 1 
The following example will illustrate the above : 

Example. A vessel floats at a draught of 16' 5^" forward, 23' ij" aft, 
the normal trim being 2 feet by the stern. At a draught of 19' 9$", her 
displacement, measured from the curve of displacement, is 5380 tons, the 
tons per inch is 31*1 tons, and the centre of flotation is 12 '9 feet abaft 
amidships. Estimate the ship's displacement. 

The difference in draught is 23' i" 1 6' 5J" = 6' 8", or 4' 8" out of 
trim. The distance between the draught-marks is 335 feet, and we 
therefore have for the thickness of the layer 

12 x 12-9 X * = 2-15 inches 

The displacement of the layer is therefore 

2-15 X 31-1 = 67 tons 
The displacement is therefore 

5380 + 67 = 5447 tons nearly 

Change of Trim due to Longitudinal Shift of 
Weights already on Board. We have seen that change 

1 This may be reduced to a formula, set as an example in Appendix A. 

T X y 
No. 2, viz. extra displacement for I foot extra trim = 12 j- , y being 

centre of flotation abaft amidships in feet. 



154 



Theoretical Naval Architecture. 



of trim is the sum of the change of draughts forwaid and aft, and 
that change of trim can be caused by the shift of weights on 
board in a fore-and-aft direction. We have here an analogous 
case to the inclining experiment in which heeling is caused by 
shifting weights in a transverse direction. In Fig. 64, let w be 




FIG. 64. 

a weight on the deck when the vessel is floating at the water- 
line WL, G being the position of the centre of gravity. Now 
suppose the weight w to be shifted forward a distance of d feet. 
G will, in consequence of this, move forward parallel to the line 
joining the original and final positions of a/, and if W be the 
displacement of the ship in tons, G will move to G' such that 

w X d 



GG' = 



W 



Now, under these circumstances, the condition of equilibrium 
is not fulfilled if the water-line remains the same, viz. that the 
centre of gravity and the centre of buoyancy must be in the 
same vertical line, because G has shifted to G'. The ship 
must therefore adjust herself till the centre of gravity and the 
centre of buoyancy are in the same vertical line, when she 
will float at a new water-line, W'L', the new centre of buoyancy 
being B'. The original vertical through G and B meets the 
new vertical through G' and B' in the point M, and this point 
will be the longitudinal metacentre, supposing the change of 
trim to be small, and GM will be the longitudinal metacentric 
height. Draw W'C parallel to the original water-line WL. 



Longitudinal Metacentre, Longitudinal BM, etc. 155 

meeting the forward perpendicular in C. Then, since CL = 
W'W, the change of trim WW'+ LL'= CL' = #, say. The 
angle of inclination of W'L' to WL is the same as the angle 
between W'L' and W'C = 6, say, and 

CL' x 

tan = ; - -r = T 
length L 

But we also have 



therefore, equating these two values for tan 6, we have 

x _GG' 
L GM 

w X d . 
"" W X GM 

using the value obtained above for GG' ; or 
*, the change of trim due to the "j ^ 

moment of transference of the > = \y x r* vf X L feet 

weight a/ through the distance </, j 
or 

. . .., 12 X w X d?X L 

The change of trim in inches = -- w v GM 

and the moment to change trim i inch is 

W x GM r 
w X d = -= foot-tons 

1 2 X \-i 

To determine this expression, we must know the vertical 
position of the centre of gravity and the position of the longi- 
tudinal metacentre. The vertical position of the centre of 
gravity will be estimated in a design when dealing with the 
metacentric height necessary, and the distance between 
the centre of buoyancy and the centre of gravity is then sub- 
tracted from the value of the longitudinal BM found by one of 
the methods already explained. The distance BG is, however, 
small compared with either of the distances BM or GM and 
any small error in estimating the position of the centre of 
gravity cannot appreciably affect the value of the moment to 
change trim one inch. In many ships BM approximately 



156 Theoretical Naval Architecture. 

equals the length of the ship, and therefore GM also ; we may 
therefore say that in such ships the moment to change trim 
i inch = -pa the displacement in tons. For ships that are long 
in proportion to the draught, the moment to change trim i inch 
is greater than would be given by this approximate rule. 

In the ship for which the value of the longitudinal BM was 
calculated on p. 148, the centre of buoyancy was 2\ feet below 
the L.W.L., the centre of gravity was estimated at i^ feet 
below the L.W.L. ; and the length between perpendiculars was 

75 f eet. 

/. GM = 69-9 - i 
= 68-9 feet 

and the moment to change trim i inch = 

12 x 75 

= 7 'oi foot-tons 
the draughts being taken at the perpendiculars. 

Example. A vessel 300 feet long and 2200 tons displacement has a 
longitudinal metacentric height of 490 feet. Find the change of trim 
caused by moving a weight of 5 tons already on board through a distance 
of 200 feet from forward to aft. 

Here the moment to change trim I inch is 

'-I^f = 300 foot-tons near l y 
The moment aft due to the shift of the weight is 

5 X 200 = looo foot-tons 
and consequently the change of trim aft is 
*ffl = 3$ inches 

Approximate Formula for the Moment to change Trim i inch. 
Assuming Normand's approximate formula for the height 
of the longitudinal metacentre above the centre of buoyancy 

given on p. 151 

A 2 X L 

H = 0-0735 -g-^ 

we may construct an approximate formula for the moment to 
change trim i inch as follows. 

We have seen that the moment to change trim i inch is 

W x GM 
12 x L 



Longitudinal Metacentre, Longitudinal BM, etc. 157 

V 
We can write W = and assume that, for our purpose, 

OJ 



BM = GM = 0- 



Substituting this in the above formula, we have 

Moment to change | _ V f A 2 x L \ 

trim i inch ) 3 5Xi2XLV '735 B x v ) 

A 2 
or 0*000175^5- 

For further approximations, see Example 18, p. 173. 
Applying this to the case worked out in detail on p. 148 

Area of L.W.P. = A = 773*5 square feet 
Breadth = B = 14-42 feet 

so that the moment to change trim i inch approximately 
should equal 

(77VO 2 

o*oooi75 v /J J/ = 7*26 foot-tons 
5 14-42 

the exact value, as calculated on p. 156, being 7*01 foot-tons. 

It is generally sufficiently accurate to assume that one-half 
the change of trim is forward, and the other half is aft. In the 
example on p. 156, if the ship floated at a draught of 12' 3" 
forward and 14' 9" aft, the new draught forward would be 

1 2' 3"- if"= 1 2' 4" 
and the new draught aft would be 

14' 9" + '?' = H' iof 

Referring, however, to Fig. 64, it will be seen that when, 
as is usually the case, the centre of flotation is not at the middle 
of the length, WW' is not equal to LL', so that, strictly speak- 
ing, the total change of trim should not be divided by 2, and 
one-half taken forward and the other half aft. Consider the 
triangles FWW, FLL'; these triangles are similar to one 



158 Theoretical Naval Architecture. 

another, and the corresponding sides are proportional, so 
that 

WW ; LL' 

~ == LF 



and both these triangles are similar to the triangle W'CL'. 
Consequently 

WW' _ LU _ CU _ change of trim 
WF ~ LF ~ W'C " length 

WF 

/. WW' = . - -T- x change of trim 
length 

T F 
and LL' = x change of trim 



that is to say, the proportion of the change of trim either aft or 
forward, is the proportion the length of the vessel abaft or 
forward of the centre of flotation bears to the length of the 
vessel. Where the change of trim is small, this makes no 
appreciable difference in the result, but there is a difference 
when large changes of trim are under consideration. 

For example, in the case worked out on p. 156, suppose 
a weight of 50 tons is moved through 100 feet from forward to 
aft ; the change of trim caused would be 

i6 inches 



The centre of flotation was 1 2 feet abaft the middle of length. 
The portion of the length abaft the centre of flotation is there- 
fore Ml f tne length. The increase of draught aft is there- 
fore 

138 v 8.0. ,-2. i nr Vips 
300* 3 13 1Ilcilc:s 

and the decrease of draught forward is 

162 y _P_ _ n i nr hpi 
300 * 3 9 "H-"Gb 

instead of 8 inches both forward and aft. The draught 
forward is therefore 

1 2' 3" -9"= n' 6" 
and the draught aft 

M' 9" + 7f = 15' 4f" 
It will be noticed that the mean draught is not the same as 



Longitudinal Metacentre, Longitudinal BM, etc. 159 

before the shifting, but two-thirds of an inch less, while the 
displacement remains the same. This is due to the fact that, 
as the ship increases her draught aft and decreases it -forward, a 
fuller portion of the ship goes into the water and a finer portion 
comes out. 

Effect on the Trim of a Ship due to adding a 
Weight of Moderate Amount. If we wish to place a 
weight on board a ship so that the vessel will not change trim, 
we must place it so that the upward force of the added buoyancy 
will act in the same line as the downward force of the added 
weight. Take a ship floating at a certain water-line, and 
imagine her to sink down a small amount, so that the new 
waterplane is parallel to the original water-plane. The added 
buoyancy is formed of a layer of parallel thickness, and having 
very nearly the shape of the original water-plane. The upward 
force of this added buoyancy will act through the centre of 
gravity of the layer, which will be very nearly vertically over 
the centre of gravity of the original water-plane, or, as we have 
termed it, the centre of flotation. We therefore see that to 
place a weight of moderate amount on a ship so that no 
change of trim takes place, we must place it vertically over or 
under the centre of flotation. The ship will then sink to a new 
water-line parallel to the original water-line, and the distance 
she will sink is known at once, if we know the tons per inch 
at the original water-line. Thus a ship is floating at a draught 
of 13 feet forward and 15 feet aft, and the tons per inch immer- 
sion is 20 tons. If a weight of 55 tons be placed over or under 
the centre of flotation, she will sink ff inches, or 2f inches, 
and the new draught will be 13' 2f" forward and 15' 2f" aft. 

It will be noticed that we have made two assumptions, both 
of which are rendered admissible by considering that the weight 
is of moderate amount. First, that the tons per inch does not 
change appreciably as the draught increases, and this is, for all 
practical purposes, the case in ordinary ships. Second, that the 
centre of gravity of the parallel layer of added buoyancy is in 
the same section as the centre of flotation. This latter assump- 
tion may be taken as true for small changes in draught caused 
by the addition of weights of moderate amount ; but for large 






160 Theoretical Naval Architecture. 

changes it will not be reasonable, because the centres of gravity 
of the water-planes are not all in the same section, but vary for 
each water-plane. As a rule, water-planes are fuller aft than 
forward near the L.W.P., and this more so as the draught 
increases ; and so, if we draw on the profile of the sheer drawing 
a curve through the centres of gravity of water-planes parallel to 
the L.W.P., we should obtain a curve which slopes somewhat 
aft as the draught increases. We shall discuss further the 
methods which have to be adopted when the weights added 
are too large for the above assumptions to be accepted. 

We see, therefore, that if we place a weight of moderate 
amount on board a ship at any other place than over the centre 
of flotation, she will not sink in the water to a water-line 
parallel to the original water-line, but she will change trim as 
well as sink bodily in the water. The change of trim will be 
forward or aft according as the weight is placed forward or 
aft of the centre of flotation. 

In determining the new draught of water, we proceed in 
two steps : 

1. Imagine the weight placed over the centre of flotation, 
and determine the consequent sinkage. 

2. Then imagine the weight shifted either forward or aft to 
the assigned position. This shift will produce a certain moment 
forward or aft, as the case may be, equal to the weight multiplied 
by its longitudinal distance from the centre of flotation. This 
moment divided by the moment to change trim i inch as cal- 
culated for the original water-plane will give the change of trim. 

The steps will be best illustrated by the following example : 

A vessel is floating at a draught of 12' 3" forward and 14' 6" aft. The 
tons per inch immersion is 2O ; length, 300 feet ; centre of flotation, 12 feet 
abaft the middle of length ; moment to change trim I inch, 300 foot-tons. 
A weight of 30 tons is placed 20 feet from the forward end of the ship. 
What will be the new draught of water ? 

The first step is to see the sinkage caused by placing the weight over 
the centre of flotation. This sinkage is i inches, and the draughts would 
then be 

12' 4$" forward, 14' 7J" aft 

Now, the shift from the centre of flotation to the given position is 142 
feet, so that the moment forward is 30 X 142 foot-tons, and the change 
of trim by the bow is 

22 4_, or 14 \ inches nearly 
300 



Longitudinal Metacentre, Longitudinal BM, etc. 161 

This has to be divided up in the ratio of 138 : 162, because the centre 
of flotation is 12 feet abaft the middle of length. We therefore have 

Increase of draught forward g X 14!" = 7f say 
Decrease of draught aft $j X 14^" = 6^" say 

The final draughts will therefore be 

Forward, 12' 4^" + 7?" = 13' oi" 
Aft, 14' 7*" - 64" = M' I" 

Effect on the Trim of a Ship due to adding a 
Weight of Considerable Amount. In this case the 
assumptions made in the previous investigation will no longer 
hold, and we must allow for the following : 

1. Variation of the tons per inch immersion as the ship 
sinks deeper in the water. 

2. The centre of flotation does not remain in the same 
transverse section. 

3. The addition of a large weight will alter the position 
of G, the centre of gravity of the ship. 

4. The different form of the volume of displacement will 
alter the position of B, the centre of buoyancy of the ship, and 
also the value of BM. 

5. Items 3 and 4 will alter the value of the moment to 
change trim i inch. 

As regards i, we can obtain first an approximation to the 
sinkage by dividing the added weight by the tons per inch 
immersion at the original water-line. The curve of tons per 
inch immersion will give the tons per inch at this new draught. 
The mean between this latter value and the original tons per 
inch, divided into the added weight, will give a very close 
approximation to the increased draught. Thus, a vessel floats at 
a constant draught of 22' 2", the tons per inch immersion 
being 44*5. It. is requiredi to find the draught after adding a 
weight of 750 tons. The first approximation to the increase of 

draught is -^ =17 inches nearly. At a draught of 23' 7" 

44'5 
it is found that the tons per inch immersion is 457. The 

mean tons per inch is therefore (44*5 + 45*7) = 45' 1 ) an d 

the increase in draught is therefore m - = 16-63, or J 6f inches 

M 



1 62 Theoretical Naval Architecture. 

nearly. This assumes that the ship sinks to a water-plane 
parallel to the first water-plane. In order that this can be the 
case, the weight must have been placed in the same transverse 
section as the centre of gravity of the layer of displacement 
between the two water-planes. We know that the weight and 
buoyancy of the ship must act in the same vertical line, and 
therefore, for the vessel to sink down without change of trim, 
the added weight must act in the same vertical line as the 
added buoyancy. We can approximate very closely to the 
centre of gravity of the layer as follows : Find the centre of 
flotation of the original W.P. and that of the parallel W.P. 
to which the vessel is supposed to sink. Put these points on 
the profile drawing at the respective water-lines. Draw a line 
joining them, and bisect this line. Then this point will be 
a very close approximation to the centre of gravity of the layer. 
A weight of 750 tons placed as above, with its centre of gravity 
in the transverse section containing 'this point, will cause the 
ship to take up a new draught of 23' 6f " with no change of trim. 
We can very readily find the new position of G, the centre 
of gravity of the ship due to the addition of the weight. Thus, 
suppose the weight of 750 tons in the above example is placed 
with its centre of gravity 16 feet below the C.G. of the ship; 
then, supposing the displacement before adding the weight to 
be 9500 tons, we have 

(750 x 16 
Lowering of G = 
10250 

= 1*17 feet 

We also have to take account of 4. In the case we have 
taken, the new C.B. below the original water-line was 97 feet, 
as against io'5 feet in the original condition, or a rise of 0*8 
foot. 

For the new water-plane we have a different longitudinal 
BM, and, knowing the new position of B and of G, we can deter- 
mine the new longitudinal metacentric height. From this we 
can obtain the new moment to change trim i inch^ using, of 
course, the new displacement. In the above case this works 
out to 950 foot-tons. 



Longitudinal Metacentre, Longitudinal BM, etc. 163 

Now we must suppose that the weight is shifted from the 
assumed position in the same vertical line as the centre of 
gravity of the layer to its given position, and this distance must 
be found. The weight multiplied by the longitudinal shift will 
give the moment changing the trim either aft or forward, as the 
case may be. Suppose, in the above case, this distance is 50 feet 
forward. Then the moment changing trim by the bow is 

750 X 50 = 37,500 foot-tons 
and the approximate change of trim is 

37,5 -T- 95 = 39* inches 

This change of trim has to be divided up in the ordinary 
way for the change of draught aft and forward. In this case 
we have 

Increase of draught forward = ~\ X 39^ = 21^ inches say 
Decrease of draught aft = Jf-f X 39^= 18 inches say 

We therefore have for our new draughts 

Draught aft, 22' 2" + i6f" - 18" =22' of" 
Draught forward, 22' 2" + i6f" + 21*" = 25' 4" 

For all ordinary purposes this would be sufficiently accu- 
rate ; but it is evidently still an approximation, because we do 
not take account of the new GM for the final water-line, and 
the consequent new moment to change trim i inch. These can 
be calculated if desired, and corrections made where necessary. 

To determine the Position of a Weight on Board 
a Ship such that the Draught aft shall remain 
constant whether the Weight is or is not on Board. 1 
Take a ship floating at the water-line WL, as in Fig. 65. If 
a weight w be placed with its centre of gravity in the transverse 
section that contains the centre of flotation, the vessel will very 
nearly sink to a parallel water-line W'L'. 2 This, however, is 
not what is required, because the draught aft is the distance 
WW greater than it should be. The weight will have to be 

1 See also Examples 25, 26 in Appendix. 

* Strictly speaking, the weight should be placed with its centre of 
gravity in the transverse section that contains the centre of gravity of the 
zone between the water-lines WL and W'L'. 



164 Theoretical Naval Architecture. 

moved forward sufficient to cause a change of trim forward of 
WW -f- LL', and then the draught aft will be the same as it 
originally was, and the draught forward will increase by the 
amount WW -f LL'. This will be more clearly seen, perhaps, 
by working the following example : 

It is desired that the draught of water aft in a steamship 
(particulars given below) shall be constant, whether the coals 




FIG. 65. 

are in or out of the ship. Find the approximate position of 
the centre of gravity of the coals in order that the desired 
condition may be fulfilled: Length of ship, 205 feet; displace- 
ment, 522 tons (no coals on board) ; centre of flotation from 
after perpendicular, 104*3 f eet ; longitudinal BM, 664 feet; 
longitudinal GM, 661*5 feet; tons per inch, 11*4; weight of 
coals, 57 tons. 

From the particulars given, we find that 

Moment to change ) 661*5 x 522 

. , } = = 140 foot-tons 

trim i inch j 12 X 205 

The bodily sinkage, supposing the coals placed with the centre 
of gravity in the transverse section containing the centre of 

flotation, will be -**- = 5 inches. Therefore the coals must 
11*4 

be shifted forward from this position through such a distance 
that a change of trim of 10 inches forward is produced. 
Accordingly, a forward moment of 

140 x 10 = 1400 foot-tons 

is required, and the distance forward of the centre of flotation 
the coals require shifting is 

= 24-6 feet 



Longitudinal Metacentre, Longitudinal BM, etc 165 
Therefore, if the coals are placed 

104*3 + 2 4'6 = 128-9 feet 

forward of the after perpendicular, the draught aft will remain 
very approximately the same as before. 

Change of Trim caused by a Compartment being 
open to the Sea. The principles involved in dealing with 
a problem of this character will be best understood by working 
out the following example : 

A rectangular-shaped lighter, 100 feet long, 40 feet broad, 
10 feet deep, floating in salt water at 3 feet level draught, has 
a collision bulkhead 6 feet from the forward end. If the side 
is broached before this bulkhead below water, what would be 
the trim in the damaged condition ? 

Let ABCD, Fig. 66, be the elevation of the lighter, with a 




PIG. 66. 

collision bulkhead 6 feet from the forward end, and floating 
at the level water-line WL. It is well to do this problem 
in two stages 

1. Determine the amount of mean sinkage due to the loss 
of buoyancy. 

2. Determine the change of trim caused. 

i. The lighter, due to the damage, loses an amount of 
buoyancy which is represented by the shaded part GB, and if 
we assume that she sinks down parallel, she will settle down at 
a water-line wl such that volume wG = volume GB. This 
will determine the distance x between wl and VVL. 






1 66 Theoretical Naval Architecture. 

For the volume wG = wH X 40 feet X x 
and the volume GB = GL x 40 feet X 3 feet 

40 x 6 x 3 18 f 
.'. x = - -if feet 

94 X 40 

= 2\ inches nearly 

2. We now deal with the change of trim caused. 
The volume of displacement = 100 x 4 X 3 cubic feet 

The weight of the lighter = I0 X 40 X 3 = 2^0 tons 

oo 

and this weight acts down through G, the centre of gravity, 
which is at 50 feet from either end. 

But we have lost the buoyancy due to the part forward of 
bulkhead EF, and the centre of buoyancy has now shifted 
back to B' such that the distance of B' from the after end is 
47 feet. Therefore we have W, the weight of lighter, acting 
down through G, and W, the upward force of buoyancy, acting 
through B'. These form a couple of magnitude 

W x 3 feet = ^P- X 3 = ^^ foot-tons 
tending to trim the ship forward. 

To find the amount of this trim, we must find the moment 
to change trim i inch 

_WxGM 
12 x L 

using the ordinary notation. 

Now, GM very nearly equals BM ; 

9400 

.". moment to change trim i inch = 7 x BM 

12 X IOO 

= X BM 



where I = the moment of inertia of the intact water-plane about 

a transverse axis through its centre of gravity ; 
and V = volume of displacement in cubic feet. 



Longitudinal Metacentre, Longitudinal BM, etc. 167 

I = ir(94 X 4) X (94)' 
V = 12000 



I440OO 

2 X 40 X (94)* 
and moment to alter trim i inch = - 

7 X 144000 

= 66 foot-tons nearly 
.*. the change of trim = ^p- -4- 66 
= 15^ inches 

The new water-line W'L' will pass through the centre 
of gravity of the water-line wl at K, and the change of trim 
aft and forward must be in the ratio 47 : 53 ; or 

Decrease of draught aft = T V^ x 15^ = 1\ inches 
Increase of draught forward == -f^ X 15^ = 8^ inches 
therefore the new draught aft is given by 
3' o" + 2j" - 7 f = 2' 7" 
and the new draught forward by 

3' o" + 2 i" + 8J" = 3' loi" 

The correctness of this result may be seen by finding the 
displacement and position of the C.B. of the new volume of 
displacement. The displacement will be found equal to the 
original displacement, and the C.B. will be found to be 50 feet 
from the after-end or the same as the C.G. 

(For a more difficult example of a similar nature, see No. 
34, Appendix A.) 



If the sums of the columns 4, 6, and 8 in the table on 
p. 148 are called Sj, S 2 , S 3 , then Area = S, X f . h, and the 
distance of the C.G. of water-plane from No. 6, viz. : 

S /z 3 

s = -?X^;I n = S 3 X2X- 
ch 3 

I the moment of inertia required. 



168 



Theoretical Naval Architecture. 



This method saves some work as compared with the above 
and is used in Brown's displacement sheet given in the Ap- 
pendix. In the above case 

7-iT (56'S6) 2 -i 

I = 2 . '- 959- 14 - v -^ 

3 L 7J 163-42 J 

This worked out by the aid of the 4 fig. logs, gives 224,300. 

To find the Longitudinal Position of the C.G. of a 
Ship. If a ship is floating at the trim assumed for the ordinary 
calculations, this is a simple matter, as the C.G. must be in 
the same vertical line as the C.B., and the longitudinal position 
of the C.B. is readily found for the draught at which the ship 
is floating. If, however, the ship is floating out of her normal 
trim, the following gives a close approximation to the position 
of the C.G. 

Suppose the ship is trimming by the stern at the water-line 
WL, as in Fig. 66A. The water-line cutting off the same 
displacement is not wl at the same mean draught as WL, but 
w'l' passing through F, the centre of flotation. The excess 




displacement over that corresponding to the mean draught is 
i2X.yxTx0, where T is the tons per inch, y the C.F. 
abaft amidships in feet, 6 is the angle between WL and wl 
(i.e. change of trim -4- length). The C.B. of the displacement 
corresponding to w'l' is at B , and can readily be determined. 
The ship, in trimming to the water-line WL, may be said to 



Longitudinal Metacentre^ Longitudinal BM, etc. 169 

pivot about a transverse axis through F, and the volume F/'L 
shifts to Ww'F. Then it can readily be shown that the stern- 
ward shift of the C.B. from B to B is BM L X 0, BM L being 
the longitudinal BM corresponding to w'l' or WL. The 
C.G. of the ship must be in the line BM L perpendicular to 
WL, and therefore G abaft mid-length = B abaft mid-length 
-(BGX0). 

If the ship trims by the bow, the C.B. shifts forward 
BM L x 0, and the C.G. is B abaft mid-length + (BG X 0). 

Change of Trim due to passing from Salt to River 
Water, or vice-versa, If the C.F. is vertically over the 
C.B. there will be no change of trim. If W is weight, and 
V and V are the volumes of displacement in salt water and 
river water (say 35 and 35-6 cubic feet to the ton respectively), 
then V = W x 35, V = W x 35'6. Let V'-V = v. 

The vessel floating at the line WL in sea water B and G 
must be in the same vertical. Supposing the ship to sink 
down parallel to W'L' in river water, the shift aft of the C.B. is 

, where b is the fore and aft separation of the C.B. and 



v . . 

C.F., and the moment changing trim = W X y , X b. The 
following example will illustrate the above. 

Example. A vessel with rectangular sections is 300 feet long, 30 feet 
broad, and floats in salt water at a draught of 15 feet forward and 20 feet aft. 
C.G. in WL. Determine the draughts forward and aft on going into 
river water 63 Ibs. to cubic foot. 
W 

Bodily sinkage =-r = 3' 34 inches. 



Volume of layer = 300 x 30 x 3-34 X T ' 5 = 2505 cubic feet. 
C.B. and .*. C.G. abaft amidships =7*15 feet. 
C.F. from amidships = nil. 

Moment to change trim one inch = 525 feet tons. 
W = 4500 tons V = 157,500 v = 2505. 

Therefore shift (forward in this case) of C.B. = x 7*15 = O'l I ft. 



The C.G. and the new C.B. are therefore o'li feet apart, and the 
moment to change trim aft is 4500 X o'li feet tons, and the change -of 
trim is I inch, say. 

The draughts are therefore 15' 3 "34" 0*5" =15' 2'84". 
,, ,, ,, 20' 3-34" +0-5" = 20' 3-84". 



17 Theoretical Naval Architecture. 

Draught of a Vessel when Launched. It is fre- 
quently necessary to make a close approximation to the 
draught forward and aft of a vessel on the occasion of launch- 
ing, and in addition to the ordinary hydrostatic curves given 
in Fig. 153 it is necessary to obtain the weight of the vessel 
on the stocks and the position of the C.G. of this weight, both 
in a longitudinal and a vertical direction. The weight will 
enable the mean draught to be fixed, taking into account the 
density of the water. At this draught the position of the 
longitudinal metacentre is known, from which the longitudinal 
GM can be found, and then the moment to change trim i inch. 
The longitudinal centre of buoyancy at the assumed draught 
can be found readily, and the moment changing trim is deter- 
mined by multiplying the weight and the longitudinal distance 
between the centre of buoyancy and the centre of gravity. 

The following example will illustrate the methods to be 
adopted 

A box-shaped vessel, 400 feet by 70 feet, floating when at 
designed draught at 22 feet forward and 24 feet aft, weighs 
before launching 6400 tons, and the position of the centre of 
gravity is i o feet abaft amidships and 3 feet below L. W.L. 
What will be her draught whe?i launched into salt water ? 

The mean draught is 8 feet, and assume she floats parallel 
to the L.W.L., 7 feet forward and 9 feet aft. At this waterline 
the C.B. is readily calculated to be 8-3 feet abaft amidships 
and 19*0 feet below L.W.L. The longitudinal BM at this 
assumed waterline works out to 1,666 feet, and the longitudinal 
GM 1650 feet, since BG is 16 feet. The moment to change 
trim i inch is 2200 foot tons. The horizontal separation of 
the C.G. and the C.B. is 10 8*3 =17 feet, so that the 

change of trim is - 2200 = 5 incnes aft - In tnis 

case, seeing that the centre of flotation is amidships, this 
5 inches is divided equally forward and aft, so that the draught 
when launched is 6 ft. 9^ in. forward and 9 ft. 2^ in. aft. 

The principal difficulty in such an estimate is the deter- 
mination of the longitudinal C.G. 



Longitudinal Metacentre, Longitudinal BM, etc. 171 

Information for use when Docking in a Floating 
Dock. When a ship is to be docked in a floating dock, 
especially if the weight is close to the lifting capacity of the 
dock, it is necessary to place the ship in the dock so that its 
C.G. is at the centre of length of the dock in order that when 
lifted the dock shall be on an even keel. 

Information in the following form is now provided to 
H.M. ships to enable the position of the C.G. to be closely 
approximated to knowing the draughts forward and aft. This 
information can be readily calculated from the sheer drawing 
by using the principles of the present chapter. 



Mean draught 
between 
draught marks. 


Corresponding 
displacement at 
normal trim. 


Longitudinal 
position of centre 
of buoyancy 
relative 
to bulkhead 132. 


Movement of the C.B. 
for every foot change 
of trim from normal 
trim by bow or stern. 


Tons 
per 
inch. 


ftt in. 


tons. 


feet. 


feet. 


tons. 


31 6 


25,900 


15*7 forward 


1-29 


*3 


28 6 


22,930 


I7'i 


1-42 


82 


256 


2O,OOO 


19-0 ,, 


F 5 8 


81 



Thus, suppose the above ship is drawing 28 ft. forward 
and 31 ft. 6 in. aft, the C.B. at the mean draught of 29 ft. 9 in. 
even keel is 16*4 ft. forward of 132, and this will move aft 
3-5 x 1*36 = 475 ft. for the 3 ft. 6 in. trim by the stern. 
The C.B. (and therefore the C.G. very nearly) is therefore 
1 6-4 4*75 = 11*65 ft- forward of 132 station, and this point 
should as nearly as possible be placed at the centre of length 
of the dock. 



Examples. (i) In the above ship if the draught forward is 25 ft. 6 in., 
and the draught aft 32 ft., estimate the position the ship should be placed 
relative to the dock. 

Ans. 7*8' forward of 132 should be well with dock centre, 
(ii) In the above ship if the draught forward is 30 ft., and aft 25 ft., 
estimate the position the ship should be placed relative to the dock. 

Ans. 25' forward of 132 should be well with dock centre. 



172 Ttieoretical Naval Architecture. 

EXAMPLES TO CHAPTER IV. 

I . A ship is floating at a draught of 20 feet forward and 22 feet aft, when 
the following weights are placed on board in the positions named : 

Weight Distance from C.G. of 

in tons. water-plane in feet. 

'}** 



}- 



What will be the new draught forward and aft, the moment to change 
trim I inch being 800 foot-tons, and the tons per inch = 35 ? 

Ans. 20' 5 f " forward, 22' 3" aft. 

2. A vessel 300 feet long, designed to float with a trim of 3 feet by 
the stern, owing to consumption of coal and stores, floats at a draught of 
9' 3" forward, and 14' 3" aft. The load displacement at a mean draught of 
13' 6" is 2140 tons ; tons per inch, i8 ; centre of flotation, 12$ feet abaft the 
middle of length. Approximate as closely as you can to the displacement. 

Ans. 1775 tons. 

3. A vessel is 300 feet long and 36 feet beam. Approximate to the 
moment to change trim I inch, the coefficient of fineness of the L.W.P. 
being 0*75. 

Ans. 319 foot-tons. 

4. A light-draught stern-wheel steamer is very approximately of the form 
of a rectangular box of 1 20 feet length and 20 feet breadth. When fully 
laden, the draught is 18 inches, and the centre of gravity of vessel and 
lading is 8 feet above the water-line. Find the transverse and longitudinal 
metacentric heights, and also the moment to change trim one inch. 

Ans. 13-47 feet, 79 1^ feet ; 56^ foot-tons. 

5. A vessel is floating at a draught of 12' 3" forward and 14' 6" aft. 
The tons per inch immersion is 20 j length, 30x3 feat ; centre of flotation, 
12 feet abaft amidships; moment to change trim I inch, 300 foot-tons. 
Where should a weight of 60 tons be placed on this vessel to bring her to 
an even keel. 

Ans. 123 feet forward of amidships. 

6. What weight placed 13 feet forward of amidships will have the same 
effect on the trim of a vessel as a weight of 5 tons placed 10 feet abaft the 
forward end, the length of the ship being 300 feet, and the centre of 
flotation 12 feet abaft amidships. 

Ans. 30 '4 tons. 

7. A right circular pontoon 50 feet long and 16 feet in diameter is just 
half immersed on an even keel. The centre of gravity is 4 feet above the 
bottom. Calculate and state in degrees the transverse heel that would be 
produced by shifting 10 tons 3 feet across the vessel. State, in inches, the 
change of trim produced by shifting 10 tons longitudinally through 20 feet. 

Ans. 3 degrees nearly ; 25 inches nearly. 

8. Show why it is that many ships floating on an even keel will increase 
the draught forward, and decrease the draught aft, or, as it is termed, go 
down by the head, if a weight is placed at the middle of the length. 

9. Show that for vessels having the ratio of the length to the draught 
about 13, the longitudinal B.M. is approximately equal to the length. 
Why should a shallow draught river steamer have a longitudinal B.M. 
much greater than the length ? What type of vessel would have a longitu- 
dinal B.M. less than the length ? 



Longitudinal Metacentre, Longitudinal BM> etc. 173 

10. Find the moment to change trim I inch of a vessel 400 feet long, 
having given the following particulars : Longitudinal metacentre above 
centre of buoyancy, 446 feet ; distance between centre of gravity and centre 
of buoyancy, 14 feet ; displacement, 15,000 tons. 

Ans. 1350 foot-tons. 

11. The moment of inertia of a water-plane of 22,500 square feet 
about a transverse axis 20 feet forward of the centre of flotation, is found 
to be 254,000,000 in foot-units. The displacement of the vessel being 
14,000 tons, determine the distance between the centre of buoyancy and 
the longitudinal metacentre. 

Ans. 500 feet. 

12. In the preceding question, if the length of the ship is 405 feet, and 
the distance between the centre of buoyancy and the centre of gravity is 13 
feet, determine the change of trim caused by the longitudinal transfer of 
150 tons through 50 feet. 

Ans. 5 1 inches nearly. 

13. A water -plane has an area of 13,200 square feet, and its moment of 
inertia about a transverse axis 14^ feet forward of its centre of gravity 
works out to 84,539,575 in foot-units. The vessel is 350 feet long, and 
has a displacement to the above water-line of 5600 tons. Determine the 
moment to change trim I inch, the distance between the centre of gravity 
and the centre of buoyancy being estimated at 8 feet 

Ans. 546 foot-tons. 

14. The semi-ordinates of a water-plane of a ship 20 feet apart are as 
follows: 0-4, 7-5, 14-5, 2i'o, 26-6, 30-9, 34-0, 36-0, 37-0, 37-3, 37-3, 
37'3. 37'3> 37'2, 37'i> 36'8, 35'8, 33'4> 28-8, 21 7, 11-5 feet respectively. 
The after appendage, whole area 214 square feet, has its centre of gravity 
6'2 feet abaft the last ordinate. Calculate 

1 i ) Area of the water-plane. 

(2) Position of C.G. of water-plane. 

(3) Transverse B.M. 

(4) Longitudinal B.M. 

(Volume of displacement up to the water-plane 525,304 cubic feet.) 
Ans. (i) 24,015 square feet; (2) 18*2 feet aoaft middle ordinate; 
(3) 17-16 feet; (4) 447-6 feet. 

15. The semi-ordinates of the L.W.P. of a vessel 15! feet apart are, 
commencing from forward, o'l, 2'5, 5-3, 8'i, io'8, 13-1, 15*0, 16*4, 17-6, 
18-3, 18-5, 18-5, 18-4, 18-1, 17-5, 16-6, 15-3, 13-3, 10-8, 7-6, 3-8 feet 
respectively. Abaft the last ordinate there is a portion of the water-plane, 
the half-area being 27 square feet, having its centre of gravity 4 feet abaft 
the last ordinate. Calculate the distance of the longitudinal metacentre 
above the centre of buoyancy, the displacement being 2206 tons. 

Ans. 534 feet. 

16. State the conditions that must hold in order that a vessel shall not 
change trim in passing from river water to salt water. 

17. A log of fir, specific gravity o'5, is 12 feet long, and the section is 
2 feet square. What is its longitudinal metacentric height when floating in 
stable equilibrium ? 

Ans. i6'5 feet nearly. 

1 8. Using the approximate formula for the moment to change trim i 
inch given on p. 157, show that this moment will be very nearly given by 

30 . -=, where T is the tons per inch immersion, and B is the breadth. 

Show also that in ships of ordinary form, the moment to change trim 
i inch approximately equals ^^ . L 2 B. 



CHAPTER V. 

STATICAL STABILITY, CURVES OF STABILITY, CALCU- 
LATIONS FOR CURVES OF STABILITY, INTEGRATOR, 
DYNAMICAL STABILITY. 

Statical Stability at Large Angles of Inclination. 
Atwood's Formula. We have up to the present only dealt 
with the stability of a ship at small angles of inclination, and 
within these limits we can determine what the statical stability is 
by using the metacentric method as explained on p. 98. We 
must now, however, investigate how the statical stability of a 
ship can be determined for large angles of inclination, because 
in service it is certain that she will be heeled over to much 
larger angles than 10 to 15, which are the limits beyond which 
we cannot employ the metacentric method. 

Let Fig. 67 represent the cross-section of a ship inclined 
to a large angle 6. WL is the position on the ship of the 
original water-line, and B the original position of the centre of 
buoyancy. In the inclined position she floats at the water-line 
W'L', which intersects WL in the point S, which for large angles 
will not usually be in the middle line of the ship. The volume 
SWW' is termed, as before, the " emerged wedge" and the volume 
SLL' the " immersed wedge" and g; g 1 are the positions of the 
centres of gravity of the emerged and immersed wedges respec- 
tively. The volume of displacement remains the same, and 
consequently these wedges are equal in volume. Let this 
volume be denoted by v. The centre of buoyancy of the 
vessel when floating at the water-line WL' is at B', and the 
upward support of the buoyancy acts through B'; the downward 
force of the weight acts through G, the centre of gravity of the 
ship. Draw GZ and BR perpendicular to the vertical through 
B', and gh,gti perpendicular to the new water-line W'L'. Then 



Statical Stability, Curves of Stability, etc. 175 

the moment of the couple tending to right the ship is W x GZ, 
or, as we term it, the moment of statical stability. Now 

GZ = BR - BP 

= BR - BG sin 

so that the moment of statical stability at the angle 6 is 
W(BR- BG.sinfl) 

The length BR is the only term in this expression that we 
do not know, and it is obtained in the following manner. The 




new volume of displacement W'AL' is obtained from the old 
volume WAL by shifting the volume WSW to the position 
LSI/, through a horizontal distance hh '. Therefore the hori- 
zontal shift of the centre of gravity of the immersed volume 
from its original position at B, or BR, is given by 

v X hK 



(using the principle discussed on p. 100). Therefore the 
moment of statical stability at the angle 6 is 

W ( V X y M - BG . sin (9 ) foot-tons 
This is known as " Atwood's formula." 



176 Theoretical Naval Architecture. 

The righting arm or lever = ^-~- BG . sin 

If G is below B, as may happen in special cases 



If we want to find the length of the righting arm or lever 
at a given angle of heel 0, we must therefore know 

(1) The position of the centre of buoyancy B in the up- 
right condition. 

(2) The position of the centre of gravity G of the ship. 

(3) The volume of displacement V. 

(4) The value of the moment of transference of the wedges 
parallel to the new water-line, viz. v X hH. 

This last expression involves a considerable amount of cal- 
culation, as the form of a ship is an irregular one. The methods 
adopted will be fully explained later, but for the present we 
will suppose that it can be obtained when the form of the ship 
is given. 

Curve of Statical Stability. The lengths of GZ thus 
obtained from Atwood's formula will vary as the angle of 
heel increases, and usually GZ gradually increases until an 
angle is reached when it obtains a maximum value. On 
further inclination, an angle will be reached when GZ becomes 
zero, and, further than this, GZ becomes negative when the 
couple W X GZ is no longer a couple tending to right the 
ship, but is an upsetting couple tending to incline the ship still 
further. Take H.M.S. Captain l as an example. The lengths 
of the lever GZ, as calculated for this ship, were as follows : 

At 1 degrees, GZ = 4^ inches At 35 degrees, GZ = 7| inches 

14 > " ~ "2 > 42 > ~ 5i 

,, 21 ,, ,, io| ,, 49 ,, ,, = 2 ,, 

,, 28 ,, = 10 ,, ,, 54 = nil 

Now set along a base-line a scale of degrees on a con- 

1 The Captain was a rigged turret-ship which foundered in the Bay of 
Biscay. A discussion of her stability will be found in "Naval Science," 
vol. i. 



Statical Stability, Curves of Stability, etc. 177 



venient scale (say \ inch = i degree), and erect ordinates 
at the above angles of the respective lengths given. If now 
we pass a curve through the tops of these ordinates, we shall 
obtain what is termed a "curve of statical stability" from 
which we can obtain the length of GZ for any angle by drawing 
the ordinate to the curve at that angle. The curve A, in Fig. 
68, is the curve so constructed for the Captain. The angle 

I 




14. 21, 2a 35. 

ANGLES OF 



42. 49. 545. 

INCLINATION. 



FIG. 68. 

at which GZ obtains its maximum value is termed the " angle of 
maximum stability" and the angle at which the curve crosses 
the base-line is termed the " angle of vanishing stability" and 
the number of degrees at which this occurs is termed the 
" range of stability" If a ship is forced over beyond the angle 
of vanishing stability, she cannot right herself; GZ having a 
negative value, the couple operating on the ship is an up- 
setting couple. 

In striking contrast to the curve of stability of the Captain 
is the curve as constructed for H.M.S. Monarch?- The lengths 
of the righting levers at different angles were calculated as 
follows : 

At 7 degrees, GZ = 4 inches 



14 

21 

28 

35 



1 The Monarch was a rigged ship built about the same time as the 
Captain^ but differing from the Captain in having greater freeboard. See 
also the volume of "Naval Science " above referred to. 



178 Theoretical Naval Architecture. 

At 42 degrees, GZ = 22 inches 



49 = 20 

" == ~ 



The curve for this ship, using the above values for GZ, is 
given by B, Fig. 68. The righting lever goes on lengthening 
in the Monarch's case up to the large angle of 40, and then 
shortens but slowly ; that of the Captain begins to shorten at 
about 21 of inclination, and disappears altogether at 54^, an 
angle at which the Monarch still possesses a large righting lever. 

Referring to Atwood's formula for the lever of statical 
stability at the angle 0, viz. 



we see that the expression consists of two parts. The 
first part is purely geometrical, depending solely upon the 
form of the ship ; the second part, BG . sin 6, brings in the 
influence of the position of the centre of gravity of the ship, 
and this depends on the distribution of the weights forming 
the structure and lading of the ship. We shall deal with these 
two parts separately. 

(1) Influence of form on curves of stability. 

(2) Influence of position of centre of gravity on curves of 
stability. 

(i) We have here to take account of the form of the ship 
above water, as well as the form of the ship below water. The 
three elements of form we shall consider are draught, beam, 
and freeboard. These are, of course, relative ; for con- 
venience we shall keep the draught constant, and see what 
variation is caused by altering the beam and freeboard. For 
the sake of simplicity, let us take floating bodies in the form 
of boxes. 1 The position of the centre of gravity is taken as 
constant. Take the standard form to be a box : 

Draught ............... 21 feet. 

Beam ............... 50^ ,, 

Freeboard ...... ... ... ... 6J ,, 



1 These illustrations are taken from a paper read at the Institution of 
Naval Architects by Sir N. Barnaby in 1871. 



Statical Stability -, Ctirves of Stability, etc. 179 

The curve of statical stability is shown in Fig. 69 by the 
curve A. The deck-edge becomes immersed at an inclination 
of 14^, and from this angle the curve increases less rapidly 
than before, and, having reached a maximum value, decreases, 
the angle of vanishing stability being reached at about 38. 

Now consider the effect of adding 4^ feet to the beam, 
thus making the box 

Draught 21 feet. 

Beam 55 

Freeboard ... ... ... ... ... 6J ,, 

The curve is now given by B, Fig. 69, the angle of vanish- 
ing stability being increased to about 45. Although the 




0. 10. 20. 30. 4O. 50. 60. 70. 

ANGLE OF INCLINATION. 

FIG. 69. 

position of the centre of gravity has remained unaltered, the 
increase of beam has caused an increase of GM, the meta- 
centric height, because the transverse metacentre has gone up. 
We know that for small angles the lever of statical stability is 
given by GM . sin 0, and consequently we should expect the 
curve B to start as shown, steeper than the curve A, because 
GM is greater. There is a very important connection between 
the metacentric height and the slope of the curve of statical 
stability at the start, to which we shall refer hereafter. 

Now consider the effect of adding 4^ feet to the freeboard 
of the original form, thus making the dimensions 

Draught 21 feet. 

Beam ... ... 50^ M 

Freeboard II 



180 Theoretical Naval Architecture. 

The curve is now given by C, Fig. 69, which is in striking 
contrast to both A and B. The angle of vanishing stability 
is now 72. The curves A and C coincide up to the angle at 
which the deck-edge of A is immersed, viz. 14^, and then, 
owing to the freeboard still being maintained, the curve C 
leaves the curve A, and does not commence to decrease 
until 40. 

These curves are very instructive in showing the influence 
of beam and freeboard on stability at large angles. We see 

(a) An increase of beam increases the initial stability, and 
therefore the slope of the curve near the origin, but does not 
greatly influence the area enclosed by the curve or the range. 

(b) An increase of freeboard has no effect on initial 
stability (supposing the increase of freeboard does not affect 
position of the centre of gravity), but has a most important 
effect in lengthening out the curve and increasing its area. 
The two bodies whose curves of statical stability are given by 
A and C have the same GM, but the curves of statical stability 
are very different. 

(2) We now have to consider the effect on the curve of 
statical stability of the position of the centre of gravity. If 
the centre of gravity G is above the centre of buoyancy B, as is 

usually the case, the righting lever is less than = by the 

expression BG . sin 6. Thus the deduction becomes greater as 
the angle of inclination increases, because sin increases as 6 
increases, reaching a maximum value of sin 6 = i when = 
90; the deduction also increases as the C.G. rises in the 
ship. Thus, suppose, in the case C above, the centre of gravity 
is raised 2 feet. Then the ordinate of the curve C at any 
angle is diminished by 2 x sin 6. For 30, sin = |, and 
the deduction is there i foot. In this way we get the curve D, 
in which the range of stability is reduced from 72 to 53 owing 
to the 2-feet rise of the centre of gravity. 

It is usual to construct these curves as indicated, the 
ordinates being righting levers, and not righting moments. The 
righting moment at any angle can be at once obtained by 
multiplying the lever by the constant displacement. The real 



Statical Stability, Ctirves of Stability, etc. 181 

curve of statical stability is of course a curve, the ordinates of 
which represent righting moments. This should not be lost 
sight of, as the following will show. In Fig. 70 are given the 



1-5.- 




75. 



curves of righting levers for a merchant vessel in two given 
conditions, A for the light condition at a displacement of 
1500 tons, and B for the load condition at a displacement of 
3500 tons. Looking simply at these curves, it would be 
thought that the ship in the light condition had the better 
stability; but in Fig. 71, in which A represents the curve of 




o. 



75. 



righting moments in the light condition, and curve B the curve 
of righting moments in the load condition, we see that the 
ship in the light condition has very much less stability than in 
the load condition. 

We see that the following are the important features of a 
curve of statical stability : 

(a) Inclination the tangent to the curve at the origin has to 
the base-line ; 

(b) The angle at which the maximum value occurs, and the 
length of the righting lever at this angle ; 

(c) The range of stability. 




1 82 Statical Stability, Curves of Stability, etc. 

The angle the tangent at the origin makes with the base- 
line can be found in a very simple manner as follows : At the 

angle whose circular measure 1 
is unity, viz. 57*3, erect a 
perpendicular to the base, 
and make its length equal to 
the metacentric height GM, 
for the condition at which 
the curve has to be drawn, 
using the same scale as for 
the righting levers (see Fig. 
72). Join the end of this 

FIG. 72. . 

line with the origin, and the 

curve as it approaches the origin will tend to lie along this line. 

The proof of this is given below. 2 

Specimen Curves of Stability. In Fig. 73 are given 
some specimen curves of stability for typical classes of ships. 

A is the curve for a modern British battleship of about 3^ 
feet metacentric height. The range is about 63. 

B is the curve for the American monitor Miantonomoh. 
This ship had a low freeboard, and to provide sufficient stability 
a very great metacentric height was provided. This is shown 
by the steepness of the curve at the start. 

C is the curve for a merchant steamer carrying a miscel- 
laneous cargo, with a metacentric height of about 2 feet. In 

1 See p. 91. 

2 For a small angle of inclination 0, we know that GZ = GM x 0, 
6 being in circular measure ; 

GZ GM 

or ~0 = -T 

If now we express in degrees, say 6 = <f>, fhen 
GZ GM 



</> angle whose circular measure is I 
G2 



Z GM 



If a is the angle OM makes with the base, then 

GM GZ 
tana = 5F3 = ^ 
and thus the line OM lies along the curve near the origin. 



Statical Stability -, Curves of Stability ', etc. 183 

this ship there is a large righting lever even at 90. It must 
be stated that, although this curve is typical for many ships, yet 




10. 20. 30. 40. 50. 60. 70. 80. 90. 

ANGLES OF INCLINATION. 



FIG. 73. 

the forms of the curves of stability for merchant steamers must 
vary considerably, owing to the many different types of ships 
and the variation in loading. Fig. 74 gives curves of stability 




10 



30 40 50 573 60 70 

DEGREES OF INCLINATION 
FIG. 74. 



for several conditions of the T.S.S. Smolensk, 470' X 58' 
X 37' (Mr. Rowell, I.N.A., 1905). They may be taken as 
typical curves of a modern steamship of the highest class. A 
is the curve for the load condition, in which the lower holds 
are filled with 1200 tons of cargo, and the 'tween decks are 
filled with 600 tons of cargo homogeneously stowed. All coal, 
stores, and water are assumed on board. The GM is 1*5 feet, 
and the range is 80. B assumes the cargo is all homo- 
geneous, the GM being reduced to i foot. The range is 
rather over 70. C is for the same cargo as B, but all coal 



1 84 



Theoretical Naval Architecture. 



is consumed except 200 tons in bottom of bunkers, and half 
stores and fresh water only remain on board. The GM is 
only o'6 foot, but the lighter draught has the effect of lengthen- 
ing out the curve to a range of yi^ . D is the condition 
when the vessel is " light," having the ballast and reserve feed 
tanks full; and the bunkers full of coal. The GM is 2 '8 feet, 
and the range is over 90. It will be noticed that the 
tangent at the origin has been drawn in each case at the angle 
f" 1 TVT 

a such that tan a = 5- 

57*3 

The curves E and F in Fig. 74A have been prepared to 
illustrate the effect of raising the centre of gravity of ship when 




10 



J 



70 



26 a ^ ___ j 

FIG. 74A. 

in condition C. If the centre of gravity is raised o'6 foot by 
a different disposition of the cargo, the GM is zero, and the 
curve of stability starts at a tangent to the base line. At all 
angles the GZ is reduced from that in condition C by o - 6 
sin 0, so that we get the curve E, in which the range is 66. 
If now we suppose the centre of gravity of ship lifted still 
higher, viz. 0-5 foot, the vessel has a negative GM in the 
upright condition, and is therefore in equilibrium, which is, 
however, unstable. This is shown by the way the curve starts 

at the origin at an angle of a = tan' 1 7-. The ship will 

heel until at 10 the centre of gravity and new CB again get 
into the same vertical line and the ship is in equilibrium. 
This time, however, the ship is in stable equilibrium, and has 
a positive GM, so that the ship will loll over to 10, and there 
be perfectly safe in calm weather, as is shown by the way the 
curve of stability stretches out to a range of 62. At sea it 



Statical Stability, Curves of Stability, etc. 185 

would be advisable to fill up some of the ballast tanks, to 
improve the stability of the vessel in either of the conditions 
E and F. 

Ships do occasionally get into the condition represented in 
F, Fig. 74A. The following has reference to the S.S. Leo, which 
capsized in 1895 (taken from Captain Owen's book "Aids to 
Stability "). This ship left port with a cargo of barley and 
wheat and 40 tons of coal on deck. Her freeboard was high, 
no ballast being taken on board, and she had a list of 10 to 
starboard or port, showing a negative GM. There was some 
loose water in the bilges which the pump suctions could not 
touch, as the ship had a list on one side or the other. The 
ship listing to starboard, the engineers, to reduce the list, used 
most of the coal from the starboard bunkers. This, however, 
had the effect of raising the centre of gravity of the ship still 
further, and increasing her instability when forced to the up- 
right. The wind and sea both now acted on the starboard 
side, so that she returned to the upright and then lurched over 
to port. The effect of the motion of the ship, and the force of 
the wind sending the ship over to leeward, caused her to go 
far beyond her natural position of equilibrium, say 10 to 15, 
and this was helped by the loose water rushing across. Conse- 
quently such an angle was reached that the shifting boards 
gave way and the grain got over to the inclined side, and the 
ship went right over. The remedy in such a case would un- 
doubtedly have been to fill the water-ballast tanks, so that the 
ship had a positive GM in the upright condition. 

A ship may start her voyage with a small positive GM, but, 
owing to consumption of coal, etc., during the voyage, she may 
get a list owing to a negative GM in the upright condition. 

The most comfortable ship at sea is one with a small GM, 
and if this is associated with such a position of the centre of 
gravity and such a freeboard that the curve gives a good 
maximum GZ and a good range, say like A in Fig. 74, we 
have most satisfactory conditions of comfort and seaworthiness. 
For small-cargo vessels it is generally recognized that the GM 
should not be less than 0*8 foot, provided that a righting arm 
of like amount is obtained at 30 to 40. 



i86 



Theoretical Naval Architecture. 



For warships the conditions of stability are special. Here, 
although the high freeboard is conducive to a good area of 
the curve of stability and a large range, as seen in Fig. 69, 
curve C, the conditions of design lead to a high position of 
the centre of gravity, because of the disposition of guns and 
armour. This discounts the effect of freeboard, as seen by 
curve D in Fig. 69. 

D, in Fig. 73, is the curve of stability for a sailing-ship 
having a metacentric height of 3^ feet. 

The curve of stability for a floating body of circular form fe 

very readily obtainable, because 
the section is such that the 
upward force of the buoyancy 
always acts through the centre 
of the section, as shown in Fig. 
75. The righting lever at any 
angle 6 is GM . sin 0, where G 
is the centre of gravity, and 
M the centre of the section. 
Taking the GM as two feet, 
then the ordinates of the curve 
of stability are o, 1*0, 173, 2-0, 
I> 73> I<0 > at intervals of 30. 

The maximum occurs at 90, and the range is 180. The 
curve is shown in Fig. 76. A similar curve is obtained for a 
submarine boat, the ordinate at angle being BG . sin 0, and 
the range 180. 



180. 



Calculations for Curves of Stability. We now pro- 
ceed to investigate methods that are or have been adopted in 
practice to determine for any given ship the curve of righting 
levers. The use of the integrator is now very general for 





Statical Stability, Curves of Stability -, etc. 187 

doing this, and it saves an enormous amount of work ; but, in 
order to get a proper grasp of the subject, it is advisable to 
understand the rrfethods that were in use previous to the intro- 
duction of the integrator. 

In constructing and using curves of stability, certain assump- 
tions have to be made. These may be stated as follows : 

1. The sides and deck are assumed to be water-tight for the 
range over which the curve is drawn. 

2. The C.G. is taken in the same position in the ship, and 
consequently we assume that no weights shift their position 
throughout the inclination. 

3. The trim is assumed to be unchanged, that is, the ship 
is supposed to be constrained to move about a horizontal longi- 
tudinal axis fixed in direction only, and to adjust herself to the 
required displacement without change of trim. 

It is not possible in this work to deal with all the systems 
of calculation that have been employed ; a selection only will 
be given in this chapter. For further information the student 
is referred to the Transactions of the Institution of Naval 
Architects, and to the work by Sir E. J. Reed on the 
11 Stability of Ships." The following are the methods that will 
be discussed : 

r. Blom's mechanical method. 

2. Barnes' method. 

3. Direct method (sometimes employed as a check on 
other methods). 

4. By Amsler's Integrator and Cross-curves of stability. 

5. Tabular method (used at Messrs. John Brown). 

6. Mr. Hok's method (given later). 

i. Blom's Mechanical Method. Take a sheet of 
drawing-paper, and prick off from the body-plan the shape of 
each equidistant section * (i.e. the ordinary sections for displace- 
ment), and cut these sections out up to the water-line at which 
the curve of' stability is required, markirg on each section the 

1 In settling the sections to be used for calculating stability by any of the 
methods, regard must be had to the existence of a poop or forecastle the 
ends of which are watertight, and the ends of these should as nearly as 
possible be made stop points in the Simpson's rule. 



1 88 Theoretical Naval Architecture. 

middle line. Now secure all these sections together in their 
proper relative positions by the smallest possible use of gum. 
The weight of these represents the displacement of the ship. 
Next cut out sections of the ship for the angle at which the 
stability is required, taking care to cut them rather above 
the real water-line, and gum together in a similar manner to 
the first set. Then balance these sections against the first 
set, and cut the sections down parallel to the inclined water- 
line until the weight equals that of the first set. When this 
is the case, we can say that at the inclined water-line the 
displacement is the same as at the original water-line in the up- 
right condition. This must, of course, be the case as the vessel 
heels over. On reference to Fig. 67, it will be seen that what 
we want to find is the line through the centre of buoyancy for 
the inclined position, perpendicular to the inclined water-line, 
so that if we can find B' for the inclined position, we can com- 
pletely determine the stability. This is done graphically by 
finding the centre of gravity of the sections we have gummed 
together, and the point thus found will give us the position of 
the centre of buoyancy fon the inclined condition. This is done 
by successively suspending the section ; and noting where the 
plumb-lines cross, as explained on p. 5 1 . Having then the centre 
of buoyancy, we can draw through it a line perpendicular to the 
inclined water-line, and if we then spot off the position of the 
centre of gravity, we can at once measure off the righting lever 
GZ. A similar set of sections must be made for each angle about 
10 apart, and thus the curve of stability can be constructed. 

2. Barnes's Method of calculating Statical 
Stability. In this method a series of tables are employed, 
called Preliminary and Combination Tables, in which the work 
is set out in tabulated form. Take the section in Fig. 77 to 
represent the ship, WL being the upright water-line for the con- 
dition at which the curve of stability is required. Now, for a 
small transverse angle of inclination it is true that the new water- 
plane for the same displacement will pass through the centre 
line of the original water-plane WL, but as the angle of inclina- 
tion increases, a plane drawn through S will cut off a volume of 
displacement sometimes greater and sometimes less than the 



Statical Stability -, Curves of Stability, etc. 189 

original volume, and the actual water-line will take up some 
such position as W'L', Fig. 77, supposing too great a volume to 
be cut off by the plane through S. Now, we cannot say 
straight off where the water-line W'L' will come. What we 
have to do is this : Assume a water-line wl passing through 
S ; find the volume of the assumed immersed wedge /SL, the 
volume of the assumed emerged wedge wSW, and the area of 
the assumed water-plane wl. Then the difference of the 
volumes of the wedges divided by the area of the water-plane 
will give the thickness of the layer between wl and the correct 
water-plane, supposing the difference of the volumes is not too 
great. If this is the case, the area of the new water-plane is 




FIG. 77. 

found, and a mean taken between it and the original. In this 
way the thickness of the layer can be correctly found. If the 
immersed wedge is in excess, the layer has to be deducted ; if 
the emerged wedge is in excess, the layer has to be added. 

To get the volumes of either of the wedges, we have to 
proceed as follows : Take radial planes a convenient angular 
interval apart, and perform for each plane the operation sym- 
bolized by ij> 2 . dx, t.e. the half-squares of the ordinates 
are put through Simpson's rule in a fore-and-aft direction for 
each of the planes. Then put the results through Simpson's 
rule, using the circular measure of the angular interval. The 



Theoretical Naval Architecture. 

result will be the volume of the wedge at the particular angle. 
For proof of this see below. 1 

' The results being obtained for the immersed and emerged 
wedges, we can now determine the thickness of the layer. This 
work is arranged as follows : The preliminary table, one table 
for each angle, consists of two parts, one for the immersed 
wedge, one for the emerged wedge. A specimen table is given 
on p. 192 for 30. The lengths of the ordinates of each radial 
plane are set down in the ordinary way, and operated on by 
Simpson's multipliers, giving us a function of the area on the 
immersed side of 550*3, and on the emerged side of 477*3. 
We then put down the squares of the ordinates, and put them 
through the Simpson's multipliers, giving us a result for the 
immersed side of 17,878, and for the emerged side 14,251. 
The remainder of the work on the preliminary table will be 
described later. 

We now proceed to the combination table for 30 (see 
p. 193), there being one table for each angle. The functions 
of squares of ordinates are put down opposite their respective 
angles, both for the immersed wedge and the emerged wedge, 
up to and including 30, and these are put through Simpson's 
multipliers. In this case the immersed wedge is in excess, and 
so we find the volume of the layer to be taken off to be 7839 
cubic feet, obtaining this by using the proper multipliers. At 
the bottom is placed the work necessary for finding the thickness 
of the layer. We have the area of the whole plane 20,540 
square feet, and this divided into the excess volume of the 
immersed wedge, 7839 cubic feet, gives the thickness of the 
layer to take off, viz. 0*382 foot, to get the true water-line. 

We now have to find the moment of transference of the 

1 The area of the section S/L is given by J ' y* . d6 y as on p. 15, and the 
volume of the wedge is found by integrating these areas right fore and 
aft, or 

!/.*.* 

which can be written 

]//>*.'* .4 

or f(lfy*.dx)a$ 

**' i fo* dx is found for each radial plane, and integrated with respect to 
the angular interval. 



Statical Stability, Curves of Stability, etc. 191 

wedges, v X M' in Atwood's formula, and this is done by using 
the assumed wedges and finding their moments about the line ST, 
and then making at the end the correction rendered necessary 
by the layer. To find these moments we proceed as follows : 
In the preliminary table are placed the cubes of the ordinates of 
the radial plane, and these are put through Simpson's rule ; the 
addition for the emerged and immersed sides are added 
together, giving us for the 30 radial plane 1,052,436. These 
sums of functions of cubes are put in the combination table for 
each radial plane up to and including 30, and they are put 
through Simpson's rule, and then respectively multiplied by 
the cosine of the angle made by each radial plane with the 
extreme radial plane at 30. The sum of these products gives 
us a function of the sum of the moments of the assumed im- 
mersed and emerged wedges about ST. The multiplier for the 
particular case given is 0*3878, so that the uncorrected moment 
of the wedges is 3,391, 336,* in foot-units, i.e. cubic feet, multi- 
plied by feet. 

1 The proof of the process is as follows : Take a section of the wedge 
S/L, Fig. 78, and draw ST perpendicular to S/. Then what is required is 
the moment of the section about ST, and this 
integrated throughout the length. Take P and 
P' on the curved boundary, very close together, 

and join SP, SP' ; call the angle P'Sl, 6, and a.vc os , 

the angle P6P', d9. -K* 2 

Then the area PS P' = %y* . de SP -y 

The centre of gravity of SPP' is distant 
from ST, \y . cos 0, and the moment of SPP' 
about ST is 

(iy > dQ ) x ^y . cos 0) FIG. 78. 

or \y* . cos . dQ 
We therefore have the moment of /SL about ST 

\jy 3 . cos B . dQ 

and therefore the moment of the wedge about ST is 
J(jJ> 3 . cos 6 . d6)dx 
or iffy 3 , cose .dx.de 

i.e. find the value of ify 3 . cos 6 . dx for radial planes up to and including 
the angle, and then integrate with respect to the angular interval. It will 
be seen that the process described above corresponds with this formula. 




PRELIMINARY TABLE FOR STABILITY. 



WATER SECTION INCLINED AT 30. 


IMMERSED WEDGE. 






-. 


g 


<_ 


** 


"8 




-. 


"S 


!i 


<5 

(4 
_C 


is* 


11 


ii 


J 


Ii 


*o 
II 


I 


Ii 


! 


B 


"g 


H 


II 


*3 


i 


3-d 

us 


1 


o S 


i 


4' 5 


j 


2'2 


20 


J 


IO 


91 


^ 


46 


2 


18-4 


2 


36-8 


339 


2 


678 


6,230 


2 


12,460 


3 


28-6 


I 


28-6 


818 


I 


818 


23,394 


I 


2 3,394 


4 


33-8 


2 


67-6 


1142 


2 


2284 


38,614 


2 


77,228 


I 


35-5 
35'6 


I 

2 


35'5 
71-2 


1260 
1267 


I 
2 


1260 
2534 


44,739 


I 
2 


44,739 
90,236 


7 


35'6 


I 


35'6 


1267 


I 


1267 


45! Il8 


I 




8 


35'6 


2 


71-2 


1267 


2 


2534 


45, IlS 


2 


90,236 


9 


35-6 


I 


35'6 


1267 


I 


1267 




I 




10 


35 -8 


2 


71-2 


1267 


2 


2534 


45!n8 


2 


90,236 


ii 




I 


33'8 


1142 


I 


1142 


38,614 


I 


38,614 


12 


26-9 


2 


53-8 


724 


2 


1448 


19,465 


2 


38,930 


13 


H'3 


* 


7-2 


205 


* 


102 


2,924 


i 


1462 








550-3 






17,878 






597,8i7 


EMERGED WEDGE. 


I 


47 


* 


2-3 


22 


J 


II 


104 


i 


52 


2 


13*8 


2 


27*6 


I9O 


2 


380 


2,628 


2 


5,256 


3 


21-4 


I 


21-4 


458 


I 


458 


9,800 


I 


9,800 


4 


27'3 


2 


54-6 


745 


2 


1490 


20,346 


2 


40,692 


5 


32-8 


I 


32-8 


1076 


I 


1076 


35,288 


I 


35,288 


6 


367 


2 


73*4 


1347 


2 


2694 


49,431 


2 


98,862 


7 


38-2 


I 


38-2 


H59 


I 


1459 


55,743 


I 


55,743 


8 




2 


73'o 


1332 


2 


2664 


48,627 


2 


97,254 


9 


33'6 


I 


33-6 


1129 


I 


1129 


37,933 


I 


37,933 


10 


29-3 


2 


58-6 


858 


2 


I7l6 


25,154 


2 




ii 


23-8 


I 


23-8 


566 


I 


5 66 


13,481 


I 


i3,'48i 


12 


16-9 


2 


33'8 


286 


2 


573 


4,827 


2 


9,654 


13 


8'4 


i 


4-2 


7i 


* 


35 


593 


i 


296 








477*3 






14,25! 


Emerged 454,619 
















Immersed 597,817 
















1,052,436 



1 The multipliers used here are half the ordinary Simpson's multipliers ; 
the results are multiplied at the end by two to allow for this. 



COMBINATION TABLE FOR STABILITY. 



CALCULATION FOR GZ AT 30. 


IMMERSED WEDGE. 


o -5 xs 




il.i 


L 


1* 


Inclinations of 
radial planes. 


'S'SP 


**! 


2 


^ 1 1 5 


Sums of functi 
of cubes of or 
nates for bot 
sides. 


Multipliers. 


Products of su 
of functions of c 
for both side 


Cosines of incl 
tions of radia 
planes. 


Functions of cu 
for moments 
wedges. 


Function 
ordinates 
radial pla 


||| 


! 


Function 
squares of 
nates for vo 
of wedg 





___ 


15,340 


i 


15,340 


974,388 


I 


974,388 


0-8660 


843,820 


IO 





15,760 


4 


63,040 


990,153 


4 


3,960,612 


0-9397 


3,721,787 


20 





16,840 


I 3 


25,260 


1,034,251 


M 


1,551,377 


0-9848 


1,527,796 


25 
30 


550 


17,701 
17,878 


2 
i 


35,402 

8,939 


1,066,771 
1,052,436 


2 


2,133,542 
526,218 


0*9962 
I -000 


2,125,434 
526,218 


Immersed wedge 147,991 






8,745,065 


Emerged wedge 134,522 


Multiplier f 


0-3878 


13,469 Uncorrected moment 


3,391,336 


Multiplier * 0-582 Correction for layer 


13,875 


Volume of layer 7,839 cub. feet. 






3,377,461 


Volume of displacement 


398,090 


Longitudinal interval = 30 feet 




BR 


8-48 


Circular measure 10 = 0*1745 


BG sin 30 = 5-95 


GZ = 2-53 


* Multiplier = X 2 X (^ X 30) X (\ X 0*1745) = 0*582 
f Multiplier = X 2 X ( X 30) X (\ X 0*1745) = 0-3878 
BG = 11-90; sin 30= 0-5 ; BG sin = 5-95 feet 


EMERGED WEDGE. 


AKEA AND POSITION OF C.G. OF RADIAL PLANE. 








15,340 


I 


15,340 


Functions Functions 
of area. of moment. 


IO 

20 
25 





14^766 
14,640 


4 

M 

2 


60,628 
22,149 
29,280 


Immersed 
Emerged 


wedge 550 
wedge 477 


17,878 
14,250 


30 


477 


14.251 


\ 


7,125 














1027 


3,628 




134,522 








Area = 


1027 X 2 X 


(i x 30) 




= 20, 540 square feet 


C.G. of radial plane on immersed side = 


1027 5 


1 77 feet 


Thickness of layer = 


7839 =Q . 


382 foot 






20540 







NOTE. The work on the preliminary table may be much simplified by 
using TchebycherFs sections : see Appendix A. 

O 



194 Theoretical Naval Architecture. 

We now have to make the correction for the layer. We 
already have the volume of the layer, and whether it has to be 
added or subtracted, and we can readily find the position of 
the centre of gravity of the radial plane. This is done at the 
bottom of the combination table from information obtained on 
the preliminary table. We assume that the centre of gravity 
of the layer is the same distance from ST as the centre of 
gravity of the radial plane, which may be taken as the case 
unless the thickness of the layer is too great. If the layer is 
thick, a new water-line is put in at thickness found, and the 
area and C.G. of this water-line found. The mean between 
the result of this and of the original plane can then be used. 
The volume of the layer, 7839 cubic feet, is multiplied by the 
distance of its centre of gravity from ST, viz. 177 feet, giving 
a result of 13,875 in foot-units, i.e. cubic feet multiplied by 
feet. The correction for the layer is added to or subtracted 
from the uncorrected moment in accordance with the following 
rules : 

If the immersed wedge is in excess, and the centre of gravity 
of the layer is on the immersed side, the correction for the layer 
has to be subtracted. 

If the immersed wedge is in excess, and the centre of gravity 
of the layer is on the emerged side, the correction for the layer 
has to be added. 

If the emerged wedge is in excess, and the centre of gravity 
of the layer is on the emerged side, the correction for the layer 
has to be subtracted. 

If the emerged wedge is in excess, and the centre of gravity 
of the layer is on the immersed side, the correction for the layer 
has to be added. 

These rules are readily proved, and are left as an exercise 
for the student. 

We, in this case, subtract the correction for the layer, 
obtaining the true moment of transference of the wedges 
as 3,377,461, or v X hh' in Atwood's formula. The volume 
of displacement is 398,090 cubic feet; BG is 11-90 feet; 
sin 30 = 0-5. So we can fill in all the items in Atwood s 
formula 



Statical Stability, Curves of Stability, etc. 195 



or GZ = 2-53 feet 

In arranging the radial planes, it has been usual to arrange 
that the deck edge comes at a stop point in Simpson's first rule, 
because there is a sudden change of ordinate as the deck edge 
is passed, and for the same reason additional intermediate 
radial planes are introduced near the deck edge. In the case 
we have been considering, the deck edge came at about 30. 
The radial planes that were used were accordingly at 

o, 10, 20, 25, 30, 35, 40, 50, 60, 70, 80, 90 

These intermediate radial planes lead to rather complicated 
Simpson's multipliers, and in order to simplify the calculations 
it is thought to be sufficiently accurate to space the radial 
planes rather closer, say every 9. For such a series of radial 
planes the multipliers for 9, 18, 27, 36, 54, 72, 81, and 90 
are readily obtained by one of Simpson's rules. For 45 the 
multipliers can be 0*4, i, i, i, I, 0*4, with the multiplier ~ x 
OT57. For 63 they can be i, f, f, f, 071, i, i, f, with the 
multiplier 0*157. These can be readily proved ; 0*157 is the 
circular measure of 9. 

Barnes's method of calculating stability has been very largely 
employed. It was introduced by Mr. F. K. Barnes at the Insti- 
tution of Naval Architects in 1861, and in 1871 a paper was 
read at the Institution by Sir W. H. White and the late Mr. 
John, giving an account of the extensions of the system, with 
specimen calculations. For further information the student is 
referred to these papers, and also to the work on " Stability," 
by Sir E. J. Reed. At the present time it is not used to any 
large extent, owing to the introduction of the integrator, 
which gives the results by a mechanical process in much less 
time. It will be seen that in using this method to find the 
stability at a given angle, we have to use all the angles up to 
and including that angle at which the stability is required. 
Thus a mistake made in the table at any of the smaller angles 
is repeated right through, and affects the accuracy of the 



1 96 



Theoretical Naval Architecture. 



calculation at the larger angles. In order to obtain an indepen- 
dent check at any required angle, we can proceed as follows : 
3. Triangular or Direct Method of calculating 
Stability. Take the body-plan, and draw on the trial plane 
through the centre of the upright water-line at the required 
angle. This may or may not cut off the required displace- 
ment. We then, by the ordinary rules of mensuration, dis- 
cussed in Chapter I., find the area of all such portions as S/L, 
Fig. 77, for all the sections, 1 and also the position of the centre 
of gravity, g, for each section, thus obtaining the distance S//. 
This is done for both the immersed and emerged wedges. The 
work can then be arranged in tabular form thus 



Number of 
section. 


Areas. 


Simpson's 
multipliers. 


Products for 
volume. 


Levers as 
&*, 


Products for 
moment about 
ST. 


I 

2 


A, 
A, 


I 
4 


A, 

4 A, 


*! 

x* 


A,*, 

4A 2 * 2 


etc. 


etc. 


etc. 


etc. 


etc. 


etc. 



S, M, 

The volume of the wedge = B! X \ common interval 
The moment of wedge about ST = Mj X \ common interval 

This being done for both wedges, and calculating the area 
of the radial plane, we can find the volume of the layer and the 
uncorrected moment of the wedges. The correction for the 
layer is added or subtracted from this, exactly as in Barnes's 
method, and the remainder of the work follows exactly the 
methods described above for Barnes's method. 

The check spot at 90 is very readily obtained, because the 
volume and C.G. of the portion above the L.W.L. are readily 
determined, and we already know the volume and C.B. of 
portion below the L.W.L. from the displacement sheet. If V 
is the volume of displacement, and Vj is one-half the volume 

V 
above the L.W.L., then - + V x is the volume when at 90 the 

ship is immersed to the middle-line plane. The C.G. of this 

1 The sections are made into simple figures, as triangles and trapeziums, 
in order to obtain the area and position of C.G. of each. 



Statical Stability, Ciirves of Stability, etc. 197 

volume is readily obtainable. The difference between this and 

V 
V will give the volume of layer, V\ = V 2 , say, where the 

layer has to be added. The C.G. of this layer is readily 
determined, as it will very nearly be that of the middle-line 
plane of the ship, so that the C.G. of the volume V is found 
at once, and this gives the GZ at 90. 

There is the disadvantage about the methods we have 
hitherto described, that we only obtain a curve of stability for 
one particular displacement, but it is often necessary to know 
the stability of a ship at very different displacements to the 
ordinary load displacement, as, for example, in the light con- 
dition, or the launching condition. The methods we are now 
about to investigate enable us to determine at once the curve 
of stability at any given displacement and any assumed position 
of the centre of gravity. 

4. Amsler's Integrator. Cross-curves of Stability. 
The Integrator is an extension of the instrument we have 
described on p. 81, known as the planimeter. A diagram of 
one form of the integrator is given in Fig. 79. A bar, BB, has 
a groove in it, and the instrument has two wheels which run in 
this groove. W is a balance weight to make the instrument 
run smoothly. There are also three small wheels that run on 
the paper, and a pointer as in the planimeter. By passing the 
pointer round an area, we can find 

(1) A number which is proportional to the area^ i.e. a 
function of the area. 

(2) A function of the moment of the area about the axis 
the bar is set to. 

(3) A function of the moment of inertia of the area about 
the same axis. 

The bar is set parallel to the axis about which moments 
are required, by means of distance pieces. 

(1) is given by the reading indicated by the wheel marked A. 

(2) is given by the reading indicated by the wheel marked M. 

(3) is given by the reading indicated by the wheel marked I. 
The finding of the moment of inertia is not required in our 

present calculation. 



I 9 8 



Theoretical Naval Architecture. 



Now let M'LMW represent the body-plan > of a vessel 
inclined to an angle of 30; then, as the instrument is set, the 




FIG. 79. 

axis of moments is the line through S perpendicular to the 
inclined water-line, and is what we have termed ST. What 
we want to find is a line through the centre of buoyancy in the 
inclined position perpendicular to the inclined water-line. By 
passing the pointer of the instrument round a section, as 
W'L'M, we can determine its area, and also its moment about 
the axis ST by using the multipliers ; and doing this for all the 
sections in the body, we can determine the displacement and 
also the moment of the displacement about ST. 2 Dividing the 
moment by the displacement, we obtain at once the distance 
of the centre of buoyancy in the inclined condition from the 
axis ST. It is convenient in practice to arrange the work in 
a similar manner to that described for the pi ani meter, p. 83, 
and the following specimen calculation for an angle of 30 will 
illustrate the method employed. 3 Every instrument has multi- 
pliers for converting the readings of the wheel A into areas, 
and those of the wheel M into moments. The multipliers 
must also take account of the scale used. 

1 The body-plan is drawn for both sides of the ship the fore-body in 
black, say, and the after-body in red. 

2 This is the simplest method, and it is the best for beginners to employ ; 
but certain modifications suggest themselves after experience with the 
instrument. See Example 23 in this chapter. 

* See Appendix A for calculation, using " TchebychefFs rule " with the 
integrator. 



Statical Stability, Ciirves of Stability, etc. 199 





AREAS. 


MOMENTS. 


SECTIONS. 


j 


I 


-ss 


d 


) 

bo 


J 


"d 


2 




1 


i 

I 


G ^ 

.5*3 


1 


1 


1 


|1 


3 




H 


Q 






M 


Q 


W g 





Initial readings 
I and 17 


3,210 


6 4 


i 


6 4 


3900 
3910 


10 


i 


10 


2,4,6,8, 10, 12, 14, and 16 
3 5> 7, 9, ii, !3> and J S 


8,859 
14,345 


5649 
5486 


4 

2 


22,596 
10,972 


3124 
2381 


+ 7*6 
+ 743 


4 

2 


3144 
1486 



33,632 4620 

Multiplier for displacement = 0*02 
Multiplier for moment = 0*2133 

Displacement = 33,632 X 0*02 = 672*6 tons 
Moment = 4620 x 0*2133 = 985 foot-tons 

GZ = 6^6= r 4 6feet 

In this case the length of the ship was divided into sixteen 
equal parts, and accordingly Simpson's first rule can be 
employed. The common interval was 8*75 feet. The multi- 
plier for the instrument was y^- for the areas, and $ for the 
moments, and, the drawing being on the scale of J inch = 
i foot, the readings for areas had to be multiplied by (4)2 = 16, 
and for moments by (4) 3 = 64. The multiplier for displace- 
ment in tons is therefore 



; 16 x ( X 8*75) x^ = 0-02 
and for the moment in foot-tons is 

li$o X 64 X (| X 8-75) X ^ = 0-2133 
We therefore have, assuming that the centre of gravity is at S 

GZ = 2 ^ = 1*46 feet 
672-6 

Now, this 672*6 tons is not the displacement up to the 
original water-line WL, and we now have to consider a new 
conception, viz. cross-curves of stability. These are the converse 



200 



Theoretical Naval Architecture. 



of the ordinary curves of stability we have been considering. 
In these we have the righting levers at a constant displacement 
and varying angles. In a cross-curve we have the righting 
levers for a constant angle, but varying displacement. Thus 
in Fig. 79, draw a water-line W"L" parallel to W'L', and for 



CROSS CURVES or STABILITY. 
C.G.IMLW.L. 




3000. 



TONS 



4000. 
DISPLACEMENT. 



SOOO 



FIG. 80. 



the volume represented by W"ML" find the displacement and 
position of the centre of buoyancy in exactly the same way as 
we have found it for the volume WML'. The distance which 
this centre of buoyancy is from the axis gives us the value of 
GZ at this displacement, supposing the centre of gravity is at 
S. The same process is gone through for two or more water- 
lines, and we shall have values of GZ at varying displacements 
at a constant angle. These can be set off as ordinates of a 
curve, the abscissae being the displacements in tons. Such a 
curve is termed the " cross-curve of stability " at 30, and for 
any intermediate displacement we can find the value of GZ at 
30 by drawing the ordinate to the curve at this displacement. 
A similar process is gone through for each angle, the same 
position for the centre of gravity being assumed all through, 



Statical Stability, Curves of Stability, etc. 201 

and a series of cross-curves obtained. Such a set of cross- 
curves is shown in Fig. 80 for displacements between 3000 
and 5000 tons at angles of 15, 30, 45, 60, 75, and 90. At 
any intermediate displacement, say at 4600 tons, we can draw 
the ordinate and measure off the values of GZ, and so obtain 
the ordi nates necessary to construct the ordinary curve of 
stability at that displacement and assumed position of the 
centre of gravity. The relation between the cross-curves and 
the ordinary curves of stability is clearly shown in Fig. 81. 




1,500 



IS. 30. 

DEGREES 



45. 60. 75. 

OF INCLINATION. 



90. 



FIG. 81. 



We have four curves of stability for a vessel at displacements 
of 1500, 2000, 2500, and 3000 tons. These are placed as 
shown in perspective. Now, through the tops of the ordinates 
at any given angle we can draw a curve, and this will be the 
cross-curve of stability at that angle. 

It will have been noticed that throughout our calculation 
we have assumed that the centre of gravity is always at the 
point S, and the position of this point should be clearly stated 
on the cross-curves. It is evident that the centre of gravity 
cannot always remain in this position, which has only been 
assumed for convenience. The correction necessary can 
readily be made as follows : If G, the centre of gravity, is 
below the assumed position S, then GZ = SZ -f SG . sin 0, and 



2O2 Theoretical Naval Architecture. 

if G is above S, then GZ = SZ - SG . sin for any angle 6. 
Thus the ordinates are measured from the cross-curves at the 
required displacement, and then, SG being known, SG sin 1 5, 
SG sin 30, etc., can be found, and the correct values of GZ 
determined for every angle. 

If an integrator is not available, cross-curves can be calcu- 
lated by using a modification of Barnes's tables already discussed. 
Three poles, O, O', O", are taken, and Barnes's tables are worked 
out for each set of radial planes through these poles ; but no 
correction is made for the layer, as for cross-curves we set off 
the lever for whatever the displacement comes to (see Example 
22). For each angle there are thus three spots obtained, and 
by the method described below tangents are drawn at each of 
these places to the curves. At 90 the result found for each 
pole should be the same; at 72 and 81, say, the spots come 
very close together and do not give reliable curves. A separate 
calculation is therefore made for 90 over a wide range of dis- 
placement, which can readily be done, and then curves for 
constant displacement are run in fair, so that auxiliary spots on 
the cross-curves at 72 and 81 are obtained. 

5. Tabular Method of calculating Stability. The 
following method of obtaining a cross-curve is a very con- 
venient one to employ ; the whole of the work being arranged 
in tabular form and Tchebycheff's rule being used, the fore 
and aft integration is easily performed by addition. Take the 
complete body of a ship, inclined as shown in Fig. Si A, the 
axis AA being perpendicular to the new water-line, and prefer- 
ably always taken through the same spot in the middle line, 
say, the intersection of the M.L. and the L.W.L. It is the 
practice at Messrs. John Brown and Co., Clydebank, to use 
this method as an independent check against the results of the 
Integrator, and the tabular form reproduced in Table IV. 1 is 
given by the kind permission of Mr. W. J. Luke. Tcheby- 
cheff s 10 ordinate rule is used for the fore and aft integration. 
In the case given, the water-lines are placed 3' 6" apart, the 
lowest one being tangential to the lowest point of the bilge. 
For each water-line the moment of the area of water-plane 

1 See at the end of the book. 



Statical Stability, Ciirves of Stability, etc. 203 

about A A is J[jy. dxj(y'f. dx\, and the area of the water-plane 
is \Jy . dx-\-jy'. dx\. Thus, for No. 4 W.L. the sum of ordinates 
port and starboard side is 296-5, and "this is a function of the 
area of the plane. As regards moment about A A, the starboard 
side sum of squares is in excess of the port side by 980, 
which is therefore a function of the moment of the plane on 
the starboard side. These functions are converted into area 
and moment, giving 15,860 square feet and 26,216 (square feet 
X feet). These are obtained in the first part of the table for 
all the water-lines considered necessary. In the second portion, 
the areas and moments of water-lines are integrated vertically 
to 2, 4, 6, 8, 10 water-lines successively as shown. The sum 
of functions of areas are turned into tons displacement ; thus, at 
2 W.L. : displacement in tons = 19,875 X 5 X 3*5 X -^ = 662 
tons. The sum of functions of moments are turned into 
moment; thus, at 2 W.L. moment in foot tons = 165,735 x \ 
X 3*5 X 3*5 = 5524. The division 5524-^-662 = 8-35 feet, 
which gives the distance of the C.B. of the displacement to 
No. 2 W.L. on the starboard side of A A, and thus one spot 
on the cross-curve at 30 is obtained, viz. 8*35 feet, at a dis- 
placement of 662 tons. If the moment on the port side is in 
excess for any W.L., that is then made negative. Thus, for 
Nos. 6, 7, 8 in the table the port side is in excess. Similarly, 
when integrating the water-lines forward and aft, if the section 
crosses the axis AA, the ordinate is given a negative value. This, 
of course, will occur more frequently as the angle gets larger. 

Tangent to a Cross-curve. If, as is usual, the cross- 
curves are drawn to represent righting levers on base of tons 
displacement, the tangent at any point on a cross-curve has an 

7 (~* >7 

inclination 6 to the base line given by tan 6 = ' - If now 
M is the righting moment in foot-tons, GZ = ^ so that 



: ,TT T l TTT ) ^y ^^y ^y 2 ~ \V\. ^W ~" W 



2O4 Theoretical Naval Architecture. 

Now, </M is the increment of moment due to increment of 

layer <AV, so that - is the distance of the C.G. of radial 

plane from ST = S^, say. Therefore tan 6 = ^(S^ - GZ). 

Sg is the same distance from ST as the centre of gravity of the 
radial plane we are dealing with, cutting off the displacement 
at which we are drawing the tangent. 

If r lt r z are ordinates of radial plane on immersed and 
emerged sides respectively, then with our usual notation 



S(r, + rjdx 

on the immersed side. These can readily be picked up on 
the Barnes's tables. If S^ > GZ, and g is to the right of S, 
then tan 6 is positive ; if S^ is to the left of S, tan is negative. 
If SG is < GZ, tan 6 is negative. These tangents must be set 
off, having in view the scales that are used for righting levers 
and for displacement. 

Dynamical Stability. The amount of work done by 
a force acting through a given distance is measured by the 
product of the force and the distance through which it acts. 
Thus, a horse exerting a pull of 30,000 Ibs. for a mile does 

30,000 X 1760 X 3 = 158,400,000 foot-lbs. of work 

Similarly, if a weight is lifted, the work done is the product of 
the weight and the distance it is lifted. In the case of a 
ship being inclined, work has to be done on the ship by some 
external forces, and it is not always possible to measure the 
work done by reference to these forces, but we can do so by 
reference to the ship herself. When the ship is at rest, we 
have seen that the vertical forces that act upon the ship are 

(1) The weight of the ship acting vertically downwards 

through the centre of gravity ; 

(2) The buoyancy acting vertically upwards through the 

centre of buoyancy ; 
these two forces being equal in magnitude. When the ship is 






Statical Stability, Curves of Stability, etc. 205 

inclined, they act throughout the whole of the inclination. 
The centre of gravity is raised, and the centre of buoyancy is 
lowered. The weight of the ship has been made to move 
upwards the distance the centre of gravity has been raised, and 
the force of the buoyancy has been made to move downwards 
the distance the centre of buoyancy has been lowered. The 
work done on the ship is equal to the weight multiplied by the 
rise of the centre of gravity added to the force of the buoyancy 
multiplied by the depression of the centre of buoyancy ; or 

Work done on the ship = weight of the ship multiplied by the 
vertical separation of the centre of gravity and the centre 
of buoyancy. 

This calculated for any given angle of inclination is termed 
" the dynamical stability " at that angle, and is the work that 
has to be expended on the ship in heeling her over to the 
given angle. 

Moseley's Formula for the Dynamical Stability 
at any Given Angle of Inclination. Let Fig. 67 repre- 
sent a vessel heeled over by some external force to the 
angle 0; g, g' being the centres of gravity of the emerged 
and immersed wedges ; gk, g'h' being drawn perpendicular to 
the new water-line W'L f . The other points in the figure have 
their usual meaning, BR and GZ being drawn perpendicular 
to the vertical through B'. 

The vertical distance between the centres of gravity and 
buoyancy when inclined at the angle 6 is B'Z. 

The original vertical distance when the vessel is upright 
is BG. 

Therefore the vertical separation is 

B'Z - BG 
and according to the definition above 

Dynamical stability = W(B'Z - BG) 
where W = the weight of the ship in tons. * 

Now, B'Z = B'R + RZ = B'R + BG . cos 
Now, using v for the volumes of either the immersed wedge 



206 Theoretical Naval Architecture. 

or the emerged wedge, and V for the volume of displace- 
ment of the ship, and using the principle given on p. 100, 
we have 



Substituting in the above value for B'Z, we have _ 



th : t sr al 

which is known as Mo s eley* s formula. 

It will be seen that this formula is very similar to Atwood's 
formula, and it is possible to calculate it out for varying 
angles by using the tables in Barnes' method of calculating 
stability. It is possible, however, to find the dynamical 
stability of a ship at any angle much more readily if the 
curve of statical stability has been constructed, and the 
method adopted, if the dynamical stability is required, is as 
follows : 



The dynamical stability of a ship at any given angle 
is equal to the area of the curve of statical 
stability up to that angle (the ordinates of this curve 
being the actual righting moments). 

Referring to Fig. 67, showing a ship heeled over to a cer- 
tain angle 0, imagine the vessel still further heeled through a 
very small additional angle, which we may call dB. The centre 
of buoyancy will move to B" (the student should here draw his 
own figure to follow the argument). B'B" will be parallel to 
the water-line W'L', and consequently the centre of buoyancy 
will not change level during the small inclination. Drawing a 
vertical B"Z' through B", we draw GZ', the new righting arm, 



Statical Stability, Ciirves of Stability, etc. 207 

perpendicular to it. Now, the angle ZGZ' = dB, and the ver- 
tical separation of Z and Z' = GZ x dB. Therefore the work 
done in inclining the ship from the angle B to the angle B -j- dd 
is 

W x (GZ . dB) 

Take now the curve of statical stability for this vessel. At 
the angle B the ordinate is GZ. Take a consecutive ordinate 
at the angle B + dQ. Then the area of such a strip = GZ x dB \ 
but this multiplied by the displacement is the same as the 
above expression for the work done in inclining the vessel 
through the angle dB t and this, being true for any small angle 
dB, is true for all the small angles up to the angle B. But the 
addition of the work done for each successive increment of 
inclination up to a given angle is the dynamical stability at 
that angle, and the sum of the areas of such strips of the curve 
of statical stability as we have dealt with above is the area of 
that curve up to the angle B. Therefore we have the dynamical 
stability of a ship at any given angle of heel is equal to the 
area of the ordinary curve of statical stability up to that angle, 
multiplied by the displacement. 

To illustrate this principle, take the case of a floating body 
whose section is in the form of a circle, and which floats with 
its centre in the surface of the water. The transverse meta- 
centre of this body must be at the centre of the circular section. 
Let the centre of gravity of the vessel be at G, and the centre 
of buoyancy be at B. Then for an inclination through 90 
G will rise till it is in the surface of the water, but the centre of 
buoyancy will always remain at the same level, so that the 
dynamical stability at 90 = W x GM. 

Now take the curve of statical stability for such a vessel. 
The ordinate of this curve at any angle = VV x GM . sin 0, 
and consequently the ordinates at angles 15 apart will be 
W . GM . sin o, W . GM . sin 15, and so on ; or 0-258 
W.GM, 0-5 W.GM, 0707 W.GM, 0-866 W . GM, 0-965 
W . GM, and W.GM. If this curve is set out, and its area 
calculated, it will be found that its area is W x GM, which is 
the same as the dynamical stability up to 90, as found above. 



2o8 Theoretical Naval Architecture. 

It should be noticed that the angular interval should not be 
taken as degrees, but should be measured in circular measure 
(see p. 90). The circular measure of 15 is 0-2618. 

The dynamical stability at any angle depends, therefore, 
on the area of the curve of statical stability up to that angle ; 
and thus we see that the area of the curve of stability is of 
importance as well as the angle at which the ship becomes un- 
stable, because it is the dynamical stability that tells us the 
work that has to be expended to force the ship over. For full 
information on this subject the student is referred to the 
" Manual of Naval Architecture," by Sir W. H. White, and Sir 
E. J. Reed's work on the " Stability of Ships." 

Mr. Hok's Method of obtaining a Curve of Sta- 
bility. In this method the ordinary planimeter is used, and 
as the use of curves through various spots obtained is a feature 
of the method, the work is readily checked as one proceeds. 
The method first obtains the curve of dynamical stability for 
a given displacement, and then from this curve the curve of 
statical stability is deduced. The former curve is the integral 
of the latter, and so the latter is the differential of the former. 
That is, if H is the dynamical stability at angle 6, then 

.GZ.</0 and GZ = ~ ~ 

Take the body prepared with the sections on both sides in 
the ordinary way inclined, as shown in Fig. SIA. Then, by 
means of the planimeter, we can determine the displacement 
up to the various dotted water-lines, and so construct a curve 
of displacement. A line parallel to the base line and distance 
away equal to the displacement V, is drawn as shown, and this 
gives the draught at which we shall cut off the displacement 
required, and for which we desire the righting lever. The 
area of owl divided by wl gives the distance of the C.B. below 
W.L., so that wb = owl-^-wl. The area owl is readily 
obtained by the planimeter. Now, G being the assumed 
position of the centre of gravity, and B' the new C.B., B being 
the C.B. when upright, the dynamical stability = W(B'Z-BG), 
and B'Z BG can be readily found by measurement. By 



Statical Stability, Curves of Stability, etc. 209 

repeating this for a number of angles, a curve of dyna- 
mical stability can be drawn, observing that the W portion 
can be left out, being constant all through. If h be such 




that H = W ./#, and ^ 10 , AJOJ ^o> etc., be the values at 10, 
20 } 3> etc -> an d GZ 10 , GZao, etc., be the values of the righting 
arm at 10, 20, etc., then at 10 we have 



o = A- (o'i745)(5 X o + 8 x 

(0*1745 . > the c.m. of 10) 

n = i(o'i745)(o + 4 X GZ^ + 



and 

By these two equations GZ 10 and GZ 20 can be obtained. Also 
Ao = f (o-i745)(o + 3 GZ 10 + 3 GZso + GZ 30 ), and GZ 10 and GZ^ 
being known GZ 30 is found. Thus values of GZ can be deter- 
mined and the ordinary curve of stability drawn in for the 
given displacement and assumed C.G. (For all practical 
purposes ^ 10 = JGZ 10 X 0*1745, seeing that the ordinary curve 
of stability is straight near the origin.) This can be done for 
other assumed displacements, and so a set of cross-curves 
drawn in for constant angles on a base of displacement, as 
already explained. All these, of course, assume a constant 
position for the C.G., and if at any displacement another 
position of the C.G. has to be allowed for, GG' sin is added 
or subtracted, according as the new G' is below or above the 
oldG. 



2IO Theoretical Naval Architecture. 

Stability of Self-righting Lifeboats. The stability of 
these boats offers several points of interest. The properties of 
such boats are 

(1) Very large watertight reserve of buoyancy, which 

renders the boat practically unsinkable. 

(2) The water shipped is automatically cleared. 

(3) The loss of stability due to shipping water is not 

sufficient to cause instability. 

(4) The boat is unstable when upside down. 

(1) For this the ends of the boat have great sheer, and the 
ends are filled in with air-cases or tanks. These cases are also 
placed under the deck and at the sides between the deck and 
seats (see Fig. SIB). The cases are in such numbers that even 
if some are broached there would still be a sufficient buoyant 
volume left. 

The large buoyant volume at the ends gives great lifting 
power to the boat when encountering a sea. 

(2) The deck of the boat is somewhat higher than the 
water-line, and in this deck, passing through to the bottom, are 
eight tubes with automatic freeing valves t which allow water to 
pass down, but not up. These are adjusted so as to drop 
down with a small pressure of water above. The rise and 
fall of the boat will soon cause any water on the deck to 
be discharged. 

The deck falls towards midships, and has a " round down," 
so that water will flow to the valves. 

(3) Shipping water on the deck has two effects on the 
stability, viz. 

(a) Will raise the C.G. due to added weight 

(b) Will make the virtual C.G. higher than the actual C.G 

with the added water, because of the free surface 

(Rise of G = where / is the moment of inertia of 

free surface, and V is volume displacement with 
added water). 

In order that these two effects may not render the boat 
unstable 

*i.) An iron keel is fitted to pull C.G. down j 



Statical Stability, Curves of Stability, etc. 



21 I 




212 



Theoretical Naval Architecture. 



(ii.) The ends remain intact, so that the lost moment of 
inertia i shall not be too great. 

The GM in this extreme condition should be positive, or 
otherwise when the boat, after capsizing, came back to the up- 
right she would again capsize, owing to the water on the deck. 

This GM will, however, only be small, and as the water- 
level falls the i will reduce considerably owing to the presence 
of the side tanks, and the water becoming less in quantity also 
helps matters. 




FIG. 8ic. 

(4) This instability when upside down is the property 
known as self-righting. In this condition, let B', M' be C.B. 
and metacentre respectively, and G the centre of gravity. 
Then G must be nearer the keel than M'. 

The buoyancy being provided largely by the ends, B' will 
be a good way from keel, G is also near keel owing to iron 
keel ; therefore B'G is large. Again, B'M' is not large, because 
the moment of inertia is provided largely by the ends ; therefore 
we get M' below G, and the boat is unstable. 

It is also necessary that the boat should have a curve of 
stability as ABC, Fig. 8ic, the only positions of equilibrium 
being at upright and 180. If the curve of stability were like 
ADFC, there would be a position of stable equilibrium at F, 
which would not be desirable. If the curve were as AEC, 
then the equilibrium at 180 would be stable. 

These boats are tested, when fully equipped, with mast up 
and sail set, by immersing to the gunwale by weights to repre- 
sent men. The boat is then turned upside down by a chain 
attached to a crane. The chain is then slipped, when the boat 
should return to the upright position. 



Statical Stability, Curves of Stability, etc. 213 

In Fig. SIB the boat is of the type with a drop keel, and 
provision is made for water ballast. There are eight automatic 
freeing valves. 

Stability of Sailing Vessels Power to carry sail. 
For comparative purposes the sail area is taken as the "plain" 
or " working " sail, and this is assumed all braced fore and 
aft. This sail for a ship would include " jib," " fore and main 
courses," " driver," three "topsails," and three "topgallant 
sails." The centre of effort vs. assumed as the C.G. of the area 
of these sails. The centre of lateral resistance of the water is 
taken as the C.G. of the middle line area. The couple heeling 
the ship is caused by the resultant of the wind pressure and 
the fluid pressure on the opposite side of the ship. If A is 
area of sails in square feet, p is pressure in Ibs. per square 
foot, h the vertical distance between the C.E. and the C.L.R., 
and 6 the angle of heel, then the moment of the couple heeling 

A y A \^s 7j 

the ship is - , and this equals the moment of stability 



W x GM X sin 0. Taking for comparative purposes a pressure 
of i Ib. per square foot (equivalent to a wind of about fourteen 
knots). 

i W x GM 
^InT = Ax h X22 *- 

This is termed the power to carry sail, and is a measure of the 
stiffness of a ship. The greater this is, the less a given vessel 
will incline under a given wind. Sailing merchant vessels 
have a value 12 to 20 ; the sloops in the Navy, 12 to 15 ; smaller 
values are usual in sailing yachts, as the crew have an influence 
on the heel by going over to the leeward side. 

If a curve of wind moment be constructed on base of 
angle and plotted on the curve of stability (ordinates repre- 
senting righting moments), where the two cross will represent 
where the stability equals the heeling moment, and this is the 
angle of steady heel in Fig. 8 ID this is 10. The area of sail 
projected on to the vertical plane is A x cos 6, and the lever 
of the moment is h x cos 0, so that the heeling moment is 
p . A . h . cos 2 6, and from this a curve can be drawn as AD, 






2I 4 



Theoretical Naval Architecture. 



Fig 8 ID. If now the ship is supposed upright and exposed 
to a sudden squall, the work done by the wind will be the area 




10 



G2O 30 4-0 

FIG. SID. 



50 



60 



under the wind moment curve. The work done to any angle 
by the stability will be the area under the curve of stability. 
At the angle of steady heel, the former is in excess by the 
amount OAB, and the ship will heel over until the area OAB 
is equal to the area BEF. If the wind remains constant, the 
ship will eventually settle to the angle of steady heel in this 
case 10. The area BEDF above the wind curve is thus the 
reserve the vessel has against further wind pressure, and this 
area is termed the " reserve of dynamical stability" 

If a ship is struck by a squall at the moment of complet- 
ing a roll to windward, say 10, the wind moment and the 
stability of the ship both act together in taking the vessel to 
the upright, the work being represented by the area OAEF, 

G 




FIG. 81 



Fig. 8 IE. It is only after passing the upright that the stability 
acts against the wind, and the ship must heel to an angle 29, 
such that the area FEABO is equal to the area BCG. It is 
thus seen that a sailing vessel with a low curve of stability, like 



Statical Stability, Curves of Stability, etc. 215 

the Captain (Fig. 68), may have insufficient reserve of dynamical 
stability, and under the above circumstances, might be blown 
right over. Mr. Wall, I.N.A., 
1914, introduced this principle 
into ordinary ship calculations. 

Heel produced by Gun 
Fire 1 (Fig. 8 IF). This problem 
has to use the principle of 
momentum. If w and W be 
the weights of projectiles (and 
powder) and the ship respec- 
tively, v the velocity of the pro- Vs -- ] ^ 

jectile, then on firing, the C.G. FIG. SIP. 

of ship will have a backward 

velocity of V, and if I is the impulsive reaction of the water 
at rather less than half draught, we have the equation 

w W 

I = .z/ - .V. 

g g 

- . v being the momentum of the shot, and .V that of the ship. 

A o 

The angular momentum of the ship is . & . , and this 

o 

has to be equated to the moment of momentum causing the 
rotation or 



from which 



V is practically negligible, so that 
w W dd 



ATJ 

-.v. AH = .k* . -r. 
g g dt 



I 72 

= *x/ 

A/ m. 



from which k z can be found. 



g 

/jf)\% 

The initial kinetic energy of the ship is J . W . k* .( -j j , and 
if 6 is the angle of heel this is equated to the dynamical 
1 See Chap. IX. on The Rolling of Ships for the definitions of T and /*. 



216 



Theoretical Naval Architecture. 



stability at 6 supposing resistances are neglected, i.e. 

J . W . m . 2 , 

regarding the curve of stability a straight line. 

v . AH\ 2 



z/.AH 



w v . AH.TT 
! W' m.g.T 



circular measure 



As an example, take a case in which 8 guns are fired on the 
broadside 25 feet above water, H being 13 feet below water, 
w = 1 100 Ibs., velocity of discharge 3000 fs., G.M. 5 feet, 
displacement 18,000 tons, T = 8 seconds. 

.. . , noo X 3000 X 38 X 8 X 180 

6 in degrees = 224O x l8>ooo x 5 x 32 - 2 x 8 

= 3j degrees nearly. 

Example. Determine the heel caused by firing simultaneously 4 guns 
at a muzzle velocity of 1600 feet per second, the weight of projectile and 
powder being 2375 Ibs., the height of guns above centre of lateral 
resistance being 30 ft., the time of a single oscillation 6 seconds, and the 
metacentric height 4^ feet. Displacement of ship 10,000 tons. 

A us. 4^ degrees. 

Angle of Heel of a Vessel when Turning. On put- 
ting a vessel's rudder over the pressure on the rudder which 

acts below the centre of lateral 
resistance tends 
vessel inwards. 




FIG. SIG. 



to heel the 
This inward 

heeling is very noticeable in the 
case of destroyers, in which the 
rudder area is relatively large. 
In ordinary ships this is only 
of short duration, and when the 
the vessel gets on the circle, an outward heel is caused by the 
centrifugal action, which acts above the centre of lateral 
resistance (C.L.R.). The moment caused by the product of 
the centrifugal force and the distance of the C.G. from the 
C.L.R. is the moment causing the heel, and this is equated to 
the moment of stability at angle 6. This is determined. 
The centrifugal force Q caused by a weight W tons moving 



Statical Stability, Curves of Stability -, etc. 217 



W z/ 2 

in a circle radius R feet at speed v feet per second is 

8 R 
W # 2 
and the moment of the couple causing heel is =? d. This 

is equated to the moment of stability at angle 6, viz. W. GM 
sin 6, or 



sn = 



= ' 88 (irds) (v in knots) - 



The features therefore which lead to a large heel when turning 
are (i) high speed, (2) small turning circle, and (3) small 
metacentric height. 

Example. A vessel whose tactical diameter is 463 yds. at a point on 
her circular turn has a speed of 15 knots, the draught being 27 ft. and the 
metacentric height 3-5 ft. Approximate to the angle of heel. 

In this case V = 15, R = 695 ft. d - 13 ft. about. GM = 3^ ft. 

/. sin 6 = 0-088 . ^ . -12 = 0-106, and 6 = 6. 

Metacentric Height when inclined about an Axis 
inclined at an Angle a with the longitudinal Middle 
Line Plane. We first have to 
find the moment of inertia of 
the waterplane about an axis 
inclined to the principal axes 
OX and OY, viz. OZ in Fig. SIR. 
O is the C.G. of waterplane. 

Drawing as in figure. 

I = 2SAxDQ 2 whereSA 
is an element of 
area 
= 2SA(j cos a x sin a) 2 




FIG. 8iH. 



cos 2 a + x 2 sin 2 a 2xy sin a cos a) 
The last term vanishes on summation since the axes are 
through the C.G. of plane. 

.-. I = cos 2 a/f . dK + sin 2 a/* 2 . dh 

I T . cos 2 a + It sin 2 a 
.'. BM tt = BM T . cos 2 a + BM L . sin 2 a 
also BG = BG . (cos 2 a + sin 2 a) since cos 2 a -f sin 2 a = i 
/. GM a = GM, . cos 2 a + GM L . sin 2 a. 



218 Theoretical Naval Architecture. 

Example. A box-shaped vessel is 80' long, 20' wide and floats at a 
draught of water of 10'. Find the value of the distance between the 
centre of buoyancy and the metacentre for inclinations about an axis 
coincident with a diagonal of the rectangular waterplane. (Honours B. of 
E. 1911). 

In this case 

a = tan- 1 \ 

sin 2 a = T ' 7 , cos 2 a = j$ 
BM T = 3'33, BMj = 53-33 
/. BM a = (3-33 X tf) + (53-33 X T ' 7 ) = 6-27 feet. 

EXAMPLES TO CHAPTER V. 

1. A two-masted cruiser of 5000 tons displacement has its centre of 
gravity at two feet above the water-line. It is decided to add a military 
top to each mast. Assuming the weight of each military top with its guns, 
men, and ready-ammunition supply to be 12 tons, with its centre of gravity 
70 feet above the water-line, what will be the effect of this change on 

(1) The metacentric height of the vessel ? 

(2) The maximum range of stability, assuming the present maximum 

range is 90, and the tangent to the curve at this point inclined 
at 45 to the base-line ? 
(Scale used, \ inch = i, \ inch = ^ foot GZ.) 

Ans. (i) Reduce 0325 foot, assuming metacentric curve horizontal ; 
(2) reduce range to about 86^, assuming no change in cross- 
curves from 5000 to 5024 tons. 

2. The curve of statical stability of a vessel has the following values of 
GZ at angular intervals of 15 : o, 0-55, i'O3, 0-99, 0*66, 0-24, and 0*20 
feet. Determine the loss in the range of stability if the C.G. of the ship 
were raised 6 inches. 

Ans. 16. 

3. Obtain, by direct application of Atwood's formula, the moment of 
stability in foot-tons at angles of 30, 60, and 90, in the case of a prismatic 
vessel 140 feet long and 40 feet square in section, when floating with sides 
vertical at a draught of 20 feet, the metacentric height being 2 feet. 

4. A body of square section of 20 feet side and 100 feet long floats with 
one face horizontal in salt water at a draught of 10 feet, the metacentric 
height being 4 inches. Find the dynamical stability at 45. 

Ans. 171 foot-tons. 

5. Indicate how far a vessel having high bulwarks is benefited by them 
as regards her stability. What precautions should be taken in their con- 
struction to prevent them becoming a source of danger rather than of safety ? 

6. Show from Atwood's formula that a ship is in stable, unstable, or 
neutral equilibrium according as the centre of gravity is below, above, or 
coincident with the transverse metacentre respectively. 

7. A vessel in a given condition displaces 4600 tons, and has the C.G. 
in the 19-feet water-line. The ordinates of the cross-curves at this 
displacement, with the C.G. assumed in the ig-feet water-line, measure as 
follows: 0*63, 1-38, 2-15, 2'o6, 1-37, 0-56 feet at angles of 15, 30, 
45, 60, 75, and 90 respectively. The metacentric height is 2-4 feet. 
Draw out the curve of stability, and state (i) the angle of maximum 
stability, (2) the angle of vanishing stability, and (3) find the dynamical 
stability at 45 and 90. 

Ans. (i) 5of : (2) iooi; (3) 3694, 9650 foot-tons. 



Statical Stability, Curves of Stability, etc. 219 

8. A vessel has a metacentric height of 3-4 feet, and the curve of stability 
has ordinates at 15, 30, 37^, 45, and 60 of 0-9, 1*92, 2*02, 1-65, and 
0*075 feet respectively. Draw out this curve, and state the angle of 
maximum stability and the angle at which the stability vanishes. 

Ans. 351, 59^- 

9. A vessel's curve of stability has the following ordinates at angles of 
15. 3o 45, 60, and 75, viz. 0-51, 0-97, 0-90, 0-53, and 0*08 feet 
respectively. Estimate the influence on the range of stability caused by 
lifting the centre of gravity of the ship o'2 foot. 

Ans. Reduce nearly 6. 

10. A square box of 18 feet side floats at a constant draught of 6 feet, 
the centre of gravity being in the water-line. Obtain, by direct drawing or 
otherwise, the value of GZ up to 90 at say 6 angles. Draw in the curve 
of statical stability, and check it by finding its area and comparing that 
with the dynamical stability of the box at 90. 

(Dynamical stability at 90 = 3 X weight of box.) 

11. A vessel fully loaded with timber, some on the upper deck, starts 
from the St. Lawrence River with a list. She has two cross-bunkers extend- 
ing to the upper deck. She reaches a British port safely, with cargo undis- 
turbed, but is now upright. State your opinion as to the cause of this. 

12. Show by reference to the curves of stability of box-shaped vessels 
on p. 174 that at the angle at which the deck edge enters the water the 
tangent to the curve makes the maximum angle with the base-line. 

13. The curve of stability of a vessel at angles of 15, 30, 45, 60, 
75, and 90 shows the following values of the righting arm, viz. 0*22, 071, 
1-05, 1-02, 0*85, and 0*56 feet respectively, the metacentric height being 
8 inches and the displacement being 4500 tons. Discuss in detail the 
condition and behaviour of the ship if 200 tons were removed from a hold 
17 feet below the centre of gravity. 

(Assume that the cross-curves from 4300 to 4500 Ions are all parallel to 
the base-line.'] 

14. A vessel of 1250 tons displacement has its centre of gravity loj feet 
above keel. The stability curve (on scale \ inch = i, \ inch ^ foot) 
ends as a straight line at 20 slope to the base line, the range being 80. 
Find the alteration in metacentric height and range of stability due to 
taking in 30 tons reserve feed water 3 feet above the keel. 

Ans. Increase 0*176 foot in GM. 
,, 5 in range. 

15. Show that for a wall-sided vessel inclined to an angle 

GZ = sin 6 (GM + JBM . tan 2 0) 
where GM and BM refer to the upright condition. 

1 6. Show, by using the above formula, that if a wall-sided vessel has a 
negative metacentric height she will loll over to an angle (J> such that 



17. Apply the answer to question 16 to show that the log shown 
floating in Fig. 55 with a negative G.M. of |, will take up the position 
corner downwards if left free. 

1 8. A box-shaped vessel is 100 feet long, 30 feet broad, and 16 feet 
deep. In the load condition the freeboard is 4 feet and the metacentric 
height is 6 feet. In the light condition the freeboard is 10 feet and the 



220 



Theoretical Naval Architecture. 



metacentric height is still 6 feet. Compare fully the stability of the vessel 
in these two conditions. (This should be treated in the light of remarks 
on p. 181.) 

19. A vessel 72 feet long floats at 6 feet draught and has 4^ feet free- 
board, with sides above water vertical. Determine the GZ at 90, the C.B. 
when upright being 3$ feet above keel, the C.G. I foot below water-line, 
the half-ordinates of water-plane being O'8, 3*3, 5*4, 6-5, 6'8, 6*3, 5*1, 2'8, 
o'6, and the displacement 100 tons. Ans. + o 67 foot. 

20. A prismatic vessel is 32 feet broad, 13 feet draught, 9 feet freeboard, 
the bilges being circular arcs of 6 feet radius. GM is 2 feet. 

(1) Obtain the first part of curve of stability by formula in question 

15 above. 

(2) Obtain values of GZ at 45 and 72 by using Barnes's method, using 

9 angular intervals. 

(3) Obtain GZ at 90 by direct method. 
Thus draw in the complete curve of stability. 

21. In the above vessel, instead of the sides above water being vertical, 
they fall in from I foot above the water-line to the deck, where the breadth 
is 24 feet. Obtain the complete curve of stability in a similar manner to 
the preceding question. 

(The curves in these two questions are given in the author's text-book 
on " Warships," chap, xix.) 

22. In obtaining cross-curves by calculation as described above, if v, 
v' are volumes of emerged and immersed wedges, between the upright 
water-plane and the radial plane through O, and^, A, g y h\ are as in our 
ordinary notation, show that the righting lever for displacement V + v' v 
is given by 



vx 



- V.OB sing 



V+z/' - v 

so that, not needing to correct for layer to get the displacement V, we get 
for example of Barnes's method in this chapter a lever of 2^52 feet at a 
displacement of 10,158 tons and angle 30. 

23. In using the integrator for stability calculation, the "figure 8" 
method is frequently employed. This consists in using the displacement 
MWL with its C.B. B as a basis, and running round sections of wedges in 
direction SL'L, WW'S. By this means the integrator adds up the 




FIG. 8xi. 



moments of wedges and subtracts the volumes. If w, itf are displacements 
of in and out wedges, the displacement to W'L' is W + (area reading X 



Statical Stability, Curves of Stability, etc. 221 

proper factor). The GZ at angle 6 and displacement W 4- w - w' is 

given by 

(moment reading X proper factor) - (W . ES . sin 6) 

{W + (area reading X proper factor)} 
24. A box-shaped vessel 420 feet long, 72 feet broad, and 24 feet 

constant draught has a compartment amidships 60 feet long, with a W.T. 

middle line bulkhead extending the whole depth. Determine the angle of 

heel caused by the vessel being bilged on one side abreast this bulkhead, 

the C.G. of the vessel being 23 feet above the keel. 

To what height should the transverse bulkheads at the ends of the 

bilged compartment be carried, so as to confine the water to this part of 

the vessel ? 

This is done in two steps : (i) sinkage, (2) heel. See Fig. 81 1. 

volume lost buoyancy 60 X 24 X 36 _ e 
Bodi* smkage = area intact w f P . = 39O x 72 = ' 8 5 feet ' 

I of intact W.P. about middlej = A . 36o . 72 , + j . fo . 36' = 12, 130, 560 

C.F. from middle line as also\ _ . f . 
the C.B. / - 



Io Ih'ough CLF: ab Ut ""I = I2 ' I3 ' 56 ~ (39 ' 72 ' lT22) 

= 12, 102, 560. 

RiSC u f bdl ? g ne - half j = 0-92 feet. 
the bodily sinkage / 

,,,. 12 . 102 . 560 , ,. . 

New B'M' =- - -=167 feet. 

420 x 72 X 24 

Vertical distance between Gj _ , fi 

and M' before heeling / ~ I2 4 07-230 

= 6*62 feet. 

The vessel must heel until G and M' are in the same vertical, so that, 
6 being the angle of heel, 

tan 6 = ?-7- = O'l6 .*. 6 = 9 nearly. 

The height of bulkhead = 24-0 + 1*85 + (37-12 X tan 0) 
= 31-8 feet nearly. 

25. Example of a similar nature for a box 400 feet by 75 feet by 26 feet, 
midship compartment with a M.L. bulkhead 50 feet long. C.G. of ship 
25 feet from keel. 

Ans. 6 = 12^, height of bulkhead, 36^23 feet. 

26. Example of a similar nature for a box 350 feet by 60 feet by 20 feet, 
midship compartment 35 feet with a M.L. bulkhead. G.M. = 8 feet. 

Ans. = 6, height of bulkhead 24*19 feet. 

27. A prismatic vessel 100 ft. long has a transverse section formed of a 
rectangle, height 10 ft. breadth 20 ft. resting on the top of a semicircle of 
radius 10 ft. The centre of gravity is 3 ft. above the keel and the draught 
of water is 10 ft. Find the volume of the correcting layer and the 
righting moment when the vessel is inclined 45, the displacement being 
constant. 

(B. of E. 1911.) 



This is an excellent example to show the application of 



222 



Theoretical Naval Architecture. 



Barnes' method, and the solution is accordingly given here- 
with. (Fig. 8ij). 

Taking angular intervals of 15 we have the following for 




FIG. 8ij. 

the first part of the combination tables. The preliminary 
tables are not necessary, seeing that the section of vessel is 
constant. 



IMMERSED WEDGE. 



EMERGED WEDGE. 



Incli- 










nation 
of 
radial 


Ordi- 

nates. 


Squares. 


S.M. 


Pro- 
ducts. 


plane. 










O 


10 


100 


I 


100 


IS 


I0'3 


1 06 


3 


318 


30 


"5 


132 


3 


396 


45 


14-14 


200 


i 


2OO 



Ordi- 

nates. 


Squares. 


S.M. 


Products. 


10 


100 


I 


IOO 


IO 


100 


3 


300 


10 


100 


3 


300 


10 


IOO 


1 


IOO 



1014 



800 



Layer = (1014 800) X ^ X f X 0*2617 
X 100 cubic ft. 

2i4X3Xo'26i7Xioo 
= - - ^ =io5ocub.ft. 

Area of radial plane = 100 X 24-14. 
Thickness of layer = 0-435 ft- 

The table for finding the uncorrected moment of the 
wedges is as follows : the cubes of the ordinates, on immersed 



Statical Stability -, Curves of Stability, etc. 223 



and emerged sides of each radial plane being added together 
before putting through the necessary multipliers to satisfy 

.cvsO.dx. dO (see p. 191). 



Angle of 
radial 
plane. 


Sums of cubes 
of ordinates 


S.M. 


Products. 


Cos0 
(from radial 
plane). 


Products. 





2,000 


r 


2,OOO 


0707 


1,414 


15 


2,093 


3 


6,279 


0-866 


5,440 


3 


2,521 


3 


7,563 


0-966 


7,300 


45 


3,828 


i 


3,828 


I'O 


3,828 



17,982 

Uncorrected moment = 17,982 X i X f X 0*2617 x 100 
= 58,750 cubic ft. x feet. 

The C.G. of the layer is on the immersed side, 2-07 ft., so that 
the layer correction is 1050 x 2*07 = 2170 cubic ft. x feet. 
The layer has to be subtracted, seeing that immersed wedge 
is in excess. The moment is accordingly deducted. The 
corrected moment is therefore 58,750 2170 = 56,580, which 
is the v X hK in At wood's formula. 

V = 100 x ~ 7 -. 100 .\ = 15,700 cubic ft. 
B below W.L. = .radius = 4-25 ft. 



GZ = 



.'. BG is 275 ft. 
yh + BG sin 0. (G below B.) 



= 3-6 -f 1-94 = 5-54 feet. 



Righting moment = 



o j 



x 5'54 = 24,800 foot tons. 



CHAPTER VI. 

CALCULATION OF WEIGHTS STRENGTH OF BUTT 
CONNECTIONS DAVITS, PILLARS, DERRICKS, AND 
SHAFT BRACKETS. 

Calculations of Weights. We have discussed in 
Chapter I. the ordinary rules of mensuration employed in find- 
ing the areas we deal with in ship calculations. For any 
given uniform plate we can at once determine the weight 
if the weight per square foot is given. For iron and steel 
plates of varying thicknesses, the weight per square foot is 
given on p. 38. For iron and steel angles and y bars of 
varying sizes and thicknesses tables are calculated, giving the 
weight per lineal foot. Such a table is given on p. 225 for 
steel angles, etc., the thicknesses being in ygths of an inch. It 
is the Admiralty practice to specify angles, bars, etc., not in thick- 
ness, but in weight per lineal foot. Thus an angle bar 3" x 3" 
is specified to weigh 7 Ibs. per lineal foot, and a Z bar 6" X 
3 i" x 3" is specified to weigh 15 Ibs. per lineal foot. When the 
bars are specified in this way, reference to tables is unnecessary. 
The same practice is employed with regard to plates, the thick- 
ness being specified as so many pounds to the square foot. 

If we have given the size of an angle bar and its thick- 
ness, we can determine its weight per foot as follows : Assume 
the bar has square corners, and is square at the root, then, if 
a and b. are the breadth of the flanges in inches, and / is the 
thickness in inches, the length of material / inches thick in the 

section is (a -f b /) inches, or feet ; and if the bar 

is of iron, the weight per lineal foot is 



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226 Theoretical Naval Architecture. 

If the bar is of steel, the weight per lineal foot is 

X 4 ' 8 X / lbs * 



Thus a 3" X 3" X f" steel angle bar would weigh 7-17 lbs., 
and a steel angle bar 3" x 3" of 7 lbs. per foot would be 
slightly less than f inch thick. 

It is frequently necessary to calculate the weight of a 
portion of a ship's structure, having given the particulars of 
its construction ; thus, for instance, a bulkhead, a deck, 
or the outer bottom plating. In any case, the first step must 
be to find the area of plating and the lengths of angle bars. 
The weight of the net area of the plating will not give us the 
total weight of the plating, because we have to allow for butt 
straps, laps, rivet-heads, and in certain cases liners. The method 
employed to find the allowance in any given case is to take a 
sample plate and find what percentage the additions come to 
that affect this plate, and to use this percentage as an addition 
to the net weight found for the whole. To illustrate this, take 
the following example : 

A deck surface of 10,335 square feet is to be covered with T 5 s -inch 
steel plating, worked flush, jointed with single-riveted edges and butts 
Find the weight of the deck, allowing 3 per cent, for rivet-heads. 

steel plates are 1275 lbs. per square foot, so that the ne 



fg-i 
ight 



Q.335 x. 275 _ 88 tons 

2240 



Now, assume an average size for the plates, say 1 6' X 4'. f-inch rivet 
will probably be used, and the width of the edge strip and butt strap wil 
be about 5 inches. The length round half the edge of the plate is 20 feet 
and the area of the strap and lap belopging to this plate is 

20 X & = 8 -33 square feet 
The percentage of the area of the plate is therefore 



X ioo = 13 per cent 
04 

Adding also 3 per cent, for rivet-heads, the total weight is 68 '4 tons. 

It is usual to add 3 per cent, to allow for the weight of rivet- 
heads. For lapped edges and butt straps, both double riveted, 



Calculation of Weights, etc. 227 

the percentage 1 comes to about 10 per cent, for laps, 5^ percent, 
for butt straps, and 3 per cent, for liners as ordinarily fitted to 
the raised strakes of plating. No definite rule can be laid 
down, because the percentage must vary according to the 
particular scantlings and method of working the plating, etc., 
specified. 

The length of stiffeners or beams required for a given area 
can be very approximately determined by dividing the area in 
square feet by the spacing of the stiffeners or beams in feet. 
For wood decks, 3 per cent, can be added for fastenings. 

Example. The beams of a deck are 3 feet apart, and weigh 22 Ibs. 
per foot run ; the deck plating weighs 10 Ibs. per square foot, and this is 
covered by teak planking 3 inches thick. Calculate the weight of a part 
54 feet long by 10 feet wide of this structure, including fastenings. 

(S. and A. Exam. 1897.) 

Net area of deck = 54 x 10 = 540 
Add for butts and laps 7 per cent. = 3 7 '8 

.577-8 
(Assume single-riveted butt straps and single-riveted laps.) 

Weight of plating = 577*8 X 10 

= 5778 Ibs. 
Running feet of beams = ^ = 180 

Weight of beams = 180 x 22 
= 3960 lbs. 

Total weight of plating and beams = 9, 738 Ibs 
Add 3 per cent, for rivet-heads = 292 ,, 

10,030 

Weight of teak 3 = 540 X % = 7155 Ibs. 
Add 3 per cent, for fastenings = 215 ,, 

Weight of wood deck 7370 ,, 



Summary. 

Plating and beams 10,030 Ibs. 

Wood deck 7,370 



Total ... 17,400 = 7'8 tons. 



1 A number of percentages worked out for various thicknesses, etc, 
will be found in Mr. Mackrow's " Pocket Book." 

2 No allowance made for beam arms, which should be done if a whole 
deck is calculated. 

a Teak taken as 53 Ibs. per cubic foot. 



228 



Theoretical Naval Architecture. 



Use of Curves. For determining the weight of some of 
the portions of a ship, the use of curves i found of very great 
assistance. Take, for instance, the transverse framing of a ship. 
For a certain length this framing will be of the same character, 
as, for example, in a battleship, within the double bottom, 
where the framing is fitted intercostally between the longi- 
tudinals. We take a convenient number of sections, say the 
sections on the sheer drawing, and calculate the weight of the 
complete frame at each section. Then along a base of length 
set up ordinates at the sections, of lengths to represent the 
calculated weights of the frames at the sections. Through the 
spots thus obtained draw a curve, which should be a fair line. 
The positions of the frames being placed on, the weight of each 
frame can be obtained by a simple measurement, and so the 
total weight of the framing determined. The curve AA in 
Fig. 82 gives a curve as constructed in this way for the transverse 
framing below armour in the double bottom of a battleship. 
Before and abaft the double bottom, where the character of the 
framing is different, curves are constructed in a similar manner. 

Weight of Outer Bottom Plating. The first step 
necessary is to determine the area we have to deal with. We 




FIG. 8a. 



can construct a curve of girths, as BB, Fig. 82 ; but the area given 
by this curve will not give us the area of the plating, becaus 
although the surface is developed in a transverse directk 



Calculation of Weights, etc. 229 

there is no development in a longitudinal direction. (Strictly 
speaking, the bottom surface of a ship is an undevelopable 
surface.) The extra area due to the slope of the level lines is 
allowed for as follows : In plate I., between stations 3 and 4, 
a line fg is drawn representing the mean slope of all the level 
lines. Then the ordinate of the curve of girths midway 
between 3 and 4 stations is increased in the ratio fg : h. This 
done all along the curve will give us a new modified curve of 
girths, as B'B', Fig. 82, and the area given by this curve will give 
a close approximation to the area of the outer bottom of the 
ship. This is, of course, a net area without allowing for butt 
straps or laps. Having a modified curve of girths for the 
whole length, we can separate it into portions over which the 
character of the plating is the same. Thus, in a vessel built 
under Lloyd's rules, the plating is of certain thickness for one- 
half the length amidships, and the thickness is reduced before 
and abaft. Also, in a battleship, the thickness of plating is the 
same for the length of the double bottom, and is reduced 
forward and aft. The curves A A and BB, Fig. 82, are con- 
structed as described above for a length of 244 feet. 

Weight of Hull. -By the use of these various methods, 
it is possible to go right through a ship and calculate the 
weight of each portion of the structure. These calculable 
portions for a battleship are 

(1) Skin-plating and plating behind armour. 

(2) Inner bottom plating. 

(3) Framing within double bottom, below armour, behind 
armour, and above armour. Outside double bottom, below 
and above the protective deck. 

(4) Steel and wood decks, platforms, beams. 

(5) Bulkheads. 

(6) Topsides. 

There are, however, a large number of items that cannot be 
directly calculated, and their weights must be estimated by 
comparison with the weights of existing ships. Such items 
are stem and stern posts, shaft brackets, engine and boiler 
bearers, rudder, pumping and ventilation arrangements, pillars, 
paint, cement, fittings, etc. 



230 Theoretical Naval Architecture. 

It is, however, a very laborious calculation to determine 
the weight of the hull of a large ship by these means; and 
more often the weight is estimated by comparison with the 
ascertained weight of existing ships. The following is one 
method of obtaining the weight of steel which would be used in 
the construction of a vessel : The size of the vessel is denoted 
by the product of the length, breadth, and depth, and for known 
ships the weight of steel is found to be a certain proportion of 
this number, the proportion varying with the type of ship. 
The coefficients thus obtained are tabulated, and for a new ship 
the weight of steel can be estimated by using a coefficient 
which has been obtained for a similar type of ship. The weight 
of wood and outfit can be estimated in a similar manner. 

Another method is described by Mr. J. Johnson, M.I.N.A., 
in the Transactions of the Institution of Naval Architects for 
1897, in which the sizes of vessels are represented by Lloyd's 
old longitttdinal number* modified as follows : In three-decked 
vessels, the girths and depths are measured to the upper deck 

1 Lloyd's numbers (now superseded by the New Rules) 

1. The scantlings and spacing of the frames, reversed frames, and floor- 
plates, and the thickness of bulkheads are regulated by nun.bers, which are 
produced as follows : 

2. For one and two decked vessels, the number is the sum of the 
measurements in feet arising from the addition of the half-moulded breadth 
of the vessel at the middle of the length, the depth from the upper part of 
the keel to the top of the upper-deck beams, with the normal round-up, 
and the girth of the half midship frame section of the vessel, measured from 
the centre line at the top of the keel to the upper-deck stringer plate. 

3. For three-deck steam -vessels, the number is produced by the 
deduction of 7 feet from the sum of the measurements taken to the top of 
the upper -deck beams. 

4. For spar-decked vessels and awning-decked steam -vessels, the 
number is the sum of the measurements in feet taken to the top of the main- 
deck beams, as described for vessels having one or two decks. 

5. The scantlings of the keel, stem, stern -frame, keelson, and stringer 
plates, the thickness of the outside plating and deck ; also the scantlings 
of the angle bars on beam stringer plates, and keelson and stringer angles 
in hold, are governed by the longitudinal number obtained by multiplying 
that which regulates the size of the frames, etc., by the length of the vessel. 

The measurements for regulating the above scantling numbers are taken 
as follows : 

I . The length is measured from the after part of the stem to the fore part 
of the stern-post on the range of the upper-deck beams in one, two, and 
three decked and spar-decked vessels, but on the range of main -deck beams 
in awning-decked vessels. 

In vessels where the stem forms a cutwater, the length is measured from 



Calculation of Weights, etc. 231 

without deducting 7 feet. In spar and awning-deck vessels, 
the girths are measured to the spar or awning decks respec- 
tively. In one, two, and well-decked vessels, the girths and 
depths are taken in the usual way. Curves are drawn for each 
type of vessel, ordinates being the weight of iron or steel in 
tons for vessels built to the highest class at Lloyd's or Veritas, 
and abscissae being Lloyd's longitudinal number modified as 
above. These curves being constructed for ships whose weights 
are known, it is a simple matter to determine the weight for a new 
ship of given dimensions. For further information the student 
is referred to the paper in volume 39 of the Transactions. 

To calculate the Position of the Centre of Gravity 
of a Ship. We have already seen in Chapter III. how to find 
the C.G. of a completed ship by means of the inclining experi- 
ment, and data obtained in this way are found very valuable in 
estimating the position of the C.G. of a ship that is being 
designed. It is evident that the C.G. of a ship when com- 
pleted should be in such a position as to obtain the metacentric 
height considered necessary, and also to cause the ship to float 
correctly at her designed trim. Suppose, in a given ship, the C.G. 
of the naked hull has been obtained from the inclining experi- 
ment (that is, the weights on board at the time of the experi- 
ment that do not form part of the hull are set down and their 
positions determined, and then the weight and position of the 
C.G. of the hull determined by the rules we have dealt with in 
Chapter III.). The position of the C.G. of hull thus determined 
is placed on the midship section, and the ratio of the distance 
of the C.G. above the top of keel to the total depth from the 
top of keel to the top of the uppermost deck amidships will 



the place where the upper-deck beam line would intersect the after edge of 
stem if it were produced in the same direction as the part below the 
cutwater. 

2. The breadth in all cases is the greatest moulded breadth of the vessel. 

3. The depth in one and two decked vessels is taken from the upper 
part of the keel to the top of the upper-deck beam at the middle of the 
length, assuming a normal round-up of beam of a quarter of an inch to a 
foot of breadth. In spar-decked vessels and awning-decked vessels, the 
depth is taken from the upper part of the keel to the top of the main-deck 
beam at the middle of the length, with the above normal round-up of 
beam. 



232 Theoretical Naval Architecture. 

give us a ratio that can be used in future ships of similar type 
for determining the position of the C.G. of the hull. Thus, in 
a certain ship the C.G. of hull was 20-3 feet above keel, the 
total depth being 34*4 feet. The above ratio in this case is 
therefore 0-59, and for a new ship of similar type, of depth 39-5 
feet, the C.G. of hull would be estimated at 39-5 Xo'59, or 23-3 
feet above the keel. For the fore-and-aft position, a similar ratio 
may be obtained between the distance of the C.G. abaft the 
middle of length and the length between perpendiculars. In- 
formation of this character tabulated for known ships is found 
of great value in rapidly estimating the position of the C.G. 
in a new design. 

For a vessel of novel type, it is, however, necessary to cal- 
culate the position of the C.G., and this is done by combining 
together all the separate portions that go to form the hull. 
Each item is dealt with separately, and its C.G. estimated as 
closely as it is possible, both vertically and in a fore-and-aft 
direction. These are put down in tabular form, and the total 
weight and position of the C.G. determined. 

In estimating the position of the C.G. of the bottom plating, 
we proceed as follows : First determine the position of the 
C.G. of the several curves forming the half-girth at the various 
stations. This is not generally at the half-girth up, but is some- 
where inside or outside the line of the curve. Fig. 83 represents 
the section AB at a certain station. The curve is divided into 
four equal parts by dividers, and the C.G. of each of these parts 
is estimated as shown. The centres of the first two portions 
are joined, and the centres of the two top portions are joined 
as shown. The centres of these last-drawn lines, g^ g^ are 
joined, and the centre of the line g^g^ viz. G, is the C.G. of 
the line forming the curve AB, and GP is the distance from the 
L.W.L. This done for each of the sections will enable us to 
put a curve, CC in Fig. 82, of distances of C.G. of the half- 
girths from the L.W.L. 1 We then proceed to find the C.G. of 
the bottom plating as indicated in the following table. The 
area is obtained by putting the half-girths (modified as already 

1 This assumes the plating of constant thickness. Plates which are 
thicker, as at keel, bilge, and sheer, can be allowed for afterwards. 



Calculation of Weights, etc. 



233 



explained) through Simpson's rule. These products are then 
multiplied in the ordinary way to find the fore-and-aft position 
of the C.G. of the plating, and also by the distances of the C.G. 




Fio. 83.* 

of the sections below the L.W.L., which distances are measured 
off from the curve CC and are placed in column 6. The 
remainder of the work does not need any further explanation. 

CALCULATION FOR AREA AND POSITION OF C.G. OF BOTTOM PLATING 
FOR A LENGTH OF 244 FEET. 



Modified 
half -girths. 


Simpson's 
multipliers. 


Products. 


Multiples 
for leverage. 


Products. 


C.G. from 
L.W.L. 


Products. 


41'5 
SI'' 


I 

4 


41-5 
204-4 


2 
I 


83-0 
204-4 


18-1 
21-6 


751 
4,415 


53'0 


2 


io6'o 


O 


287-4 


22'2 


2,354 


49 -6 
37'5 


4 


198-4 
37'5 


2 


198-4 

75'o 


21-3 
I9-0 


4,226 
712 



587-8 



273-4 

14-0 



12,458 



1 The C.G. of wood sheathing, if fitted, can be obtained from this 
figure by setting off normally to the curve from G the half-thickness of the 
sheathing. 



234 



Theoretical Naval Architecture. 



Common interval = 61 feet 
Area both sides = 587*8 x \ X 61 x 2 
= 23,904 square feet 

C.G. abaft middle of length of plating = - -= x 61 

557-5 

= 1*45 feet 

C.G. below L.W.L. = Q >45 * = 21-2 feet 
587-8 

CALCULATION FOR THE POSITION OF THE C.G. OF A VESSEL. 







FROM L.W.L. 


FROM MIDDLE OF LENGTH. 


IM.BI 


| 


Below. 


Above. 


Before. 


Abaft. 


1TBMB. 


s 




_. 




. 




J 




. 






i 


| 


t> 


| 


B 


5 


w 


s 








1 


3 







% 


* 


* 


Equipment 




















Water 


25 


4-0 


100 














I2'0 


300 


Provisions 


3 


4-5 


135 







25-0 


75o 







Officers' stores 


15 


2'0 


30 











I25-0 


i,875 


Officers,men,and effects 


30 







6-0 


i so 


55-01650 







Cables 


30 


4-0 


120 






85-02550 








Anchors 


10 







15*0 


150 


90-0 


900 








Masts, yards, etc. 


25 








45-0 


1125 








7'0 


175 


Boats 


IO 








21'0 


210 








20'0 


200 


Warrant officers' stores 


20 


1-5 


30 








65-0 


1300 








Armament 


*75 







4 


700 







5*o 


875 


Machinery 


450 


4-0 


1800 















14,850 


Engineers' stores 


So 


'S 


25 














70-0 


3,500 


Coals 


300 


0'2 


60 








3*0 


900 







Protective deck 


2IO 








i -5 


315 







15-0 


3,150 


Hull 


1250 


__ 


"^ 


i-5 


1875 


~ 


^~~ 


"'5 


14,375 


Total 2630 2300 4555 8050 39,300 
tons 2300 8,050 



2630)2255 



0-86 ft. 
above L.W.L 



C.G. above L.W.L. = O'86 
Trans, met. ,, =2-97 

Trans, met. above C.G. = 2*97 0*86 
= 2 ' 1 1 feet. 



2630)31,250 

1 1 -88 ft. 
abaft mid. 
length. 



Calculation of Weights, etc. 2 35 

Calculation for C.G. of a Completed Vessel. By the 
use of the foregoing methods we can arrive at an estimate of 
the weight of hull, and also of the position of its C.G. relative 
to a horizontal plane, as the L.W.P., and to a vertical athwart- 
ship plane, as the midship section. To complete the ship for 
service, there have to be added the equipment, machinery, etc., 
and the weights of these are estimated, as also the positions of 
their centres of gravity. The whole is then combined in a 
table, and the position of the C.G. of the ship in the completed 
condition determined. 

The preceding is such a table as would be prepared for a 
small protected cruiser. It should be stated that the table is 
not intended to represent any special ship, but only the type of 
calculation. 

The total weight is 2630 tons, and the C.G. is 0*86 foot 
above the L.W.L. and 11*88 feet abaft the middle of length. 
The sheer drawing enables us to determine the position of the 
transverse metacentre, and the estimated GM. is found to be 
2'ii feet. The centre of buoyancy calculated from the sheer 
drawing should also be, if the ship is to trim correctly, at a 
distance of 1 1 '88 feet abaft the middle of length. 

Strength of Butt Fastenings. Fig. 84 represents two 
plates connected together by an ordinary treble-riveted butt 
strap. The spacing of the rivets in the line of holes nearest 
the butt is such that the joint can be caulked and made water- 
tight, and the alternate rivets are left out of the row of holes 
farthest from the butt. Such a connection as this could con- 
ceivably break in five distinct ways 

1. By the whole of the rivets on one side of the butt 
shearing. 

2. By the plate breaking through the line of holes, AA, 
farthest from the butt. 

3. By the butt strap breaking through the line of holes, BB, 
nearest the butt. 

4. By the plate breaking through the middle row of holes, 
CC, and shearing the rivets in the line A A. 

5. By the strap breaking through the middle row of holes, 
CC, and shearing the rivets in the line BB. 



2 3 6 



Theoretical Naval Architecture. 



It is impossible to make such a connection as this equal to 
the strength of the unpunched plate, because, although we might 



i 


*JD.B 










ffa t 


ipi 


f: 


,a i< 


>! 




i 


4 > 


t 


I . { 


, 




fT( 


] 






r , 




1ft 


4 H 


"!M 


, ^ !< 


- 




; i i< 












if 


i.,M 


r 




. 




/ if! %? 





>4* 


! 






( Li 

* ii 
)i 


f, 




j< 


^ 




t 


j^ 


* -4 M 


i h 






* * 


1 










[i ii 

i-4-l < 


1 


'f 


i '[ 






tr 


A.Ci 


I. 




i 

,_ 





FIG. 84. 



put in a larger number of rivets and thicken up the butt stiap, 




FIG. 85. 



there would still remain the line of weakness of the plate 
through the line of holes, AA, farthest from the butt. 



Calculation of Weights, etc. 237 

The most efficient form of strap to connect two plates 
together would be as shown in Fig. 85, of diamond shape. 
Here the plate is only weakened to the extent of one rivet-hole. 
Such an efficient connection as this is not required in ship con- 
struction, because in all the plating we have to deal with, such 
as stringers and outer bottom-plating, the plate is necessarily 
weakened by the holes required for its connection to the 
beam or frame, and it is unnecessary to make the connection 
stronger than the plate is at a line of holes for connecting it to 
the beam or frame. In calculating the strength of a butt con- 
nection, therefore, we take as the standard strength the strength 
through the line of holes at a beam or frame, and we so 
arrange the butt strap that the strength by any of the modes 
of fracture will at least equal this standard strength. 

Experimental Data. Before we can proceed to calcu- 
late the strength of these butt connections, we must have some 
experimental data as to the tensile strength of plates and the 
shearing strength of rivets. The results of a series of experi- 
ments were given by Mr. J. G. Wildish at the Institution of 
Naval Architects in 1885, and the following are some of the 
results given : 

SHEARING STRENGTH OF RIVETS IN TONS. 
(Pan heads and countersunk points.) 

Single shear. Double shear. 

\ inch iron rivets in iron plates ... 10*0 18 

3 ,, steel 8-4 

\ inch steel ,, ,, ,, ii'S 21'2 

i ,, ,, I5-25 

i'o ,, ,, ,, ,, 20-25 

It will be noticed that the shearing strength of the steel 
rivets of varying sizes is very nearly proportionate to the sec- 
tional area of the rivets. Taking the shearing strength of a 
f-inch steel rivet to be 11*5 tons, the strength proportionate to 
the area would be for a J-inch steel rivet, 15 '6 tons, and for a 
i-inch steel rivet 20*4 tons. Also, we see that the double shear 
of a rivet is about 1-8 times the single shear. 

The following results were given as the results of tests of 
mild steel plates : 



Theoretical Naval Architecture. 



Unpunched 28J tons per square inch. 

Holes punched 22 ,, ,, 

or a depreciation of 22 per cent. 

Holes drilled 29^ tons per square inch. 

Holes punc /W small, and the hole then \ 
countersunk / 



29 



The following give the strength of the material of the plates 
after being connected together by a butt strap : 



24-9 tons per square inch. 

Holes punched small and then countersunk, I 

the rivets being panhead, with countersunk > 28*9 ,, 

points 

It appears, from the above results, that if a plate has the 
holes drilled or has them punched and countersunk in the 
ordinary way as for flush riveting, the strength of the material 
is fully maintained. Also that, although punching holes in a 
plate reduces the strength from 28^ to 22 tons per square inch, 
a reduction of 22 per cent., yet when connected by a butt strap, 
and riveted up, the strength rises to 24-9 tons per square inch, 
which is only 12 per cent, weaker than the unpunched plate, 
the process of riveting strengthening the plate. 

The following table giving the results of more recent experi- 
ments, go to confirm the above figures : 

TABLE OF BREAKING STRESS IN TONS PER SQUARE INCH (OF AREA 
OF PLATING BEFORE TESTING). 



Nature of test. 


Mild 


steel. 


High tensile 
steel. 






(2) 28*0 




Elongation % in 8 in 


(15) 25 % 

(8) 2Q'6 


(2) 29-8 % 
(2) 30'6 


(5) 24-7 % 






(8) 3i '6 




Plate punched not countersunk, unriveted 
ii n ii riveted . 
,, ,, countersunk and riveted . 


(16} 26-5 
(33) 29*6 

(16) 29-6 


22'I 

247 
28-2 


(6) 267 
(12} 25-2 
(12) 29-0 . 


Single 5 in. rivet, counter--* Mnd , . 
sunk point, tons per| H T rfvet 

Double.shear in. rivet, \ M; , , , 
countersunk point, tons 1 Mild steel rivet 
per rivet 


O) i4'8 
(2) 217 

(2) 39'7 


*5'7 

28-9 


(2) 21-0 



Figures in brackets ( ) denote the number of tests of which the result given is the mean. 



In an ordinary butt-strap, with the holes spaced closely 



Calculation of Weights, etc. 239 

together in order to obtain a water-tight pitch for the rivets, it 
is found that the punching distresses the material in the neigh- 
bourhood of the holes, and the strength is materially reduced, 
as we have seen above, but after riveting the strength is to 
some extent restored. It was formerly the practice to anneal 
butt straps of steel plating, but this practice is now discontinued 
in both Admiralty and Lloyd's practice. 

In our calculations of the strength of butt straps, we assume 
that the strength of the material between the rivet-holes is the 
same as the strength of the material of the unpunched plate. 

Again, the plating, in the cases we have to deal with, has 
the riveting flush on the outside, and the holes are made with 
a countersink for this purpose. Here also we can assume that 
the strength of the material is the same as the strength of the 
material of the unpunched plate. 

The specified tests for the tensile strength of mild steel 
plates are as follows : 

For ships built for the British Admiralty, not less than 26 
and not more than 30 tons per square inch of section. 

For ships built to the rules of Lloyd's Register, not less 
than 28 and not more than 32 tons per square inch of 
section. 

The plates tested above showed a tensile strength of about 
28 tons per square inch, or nearly midway between the limits 
laid down by the British Admiralty. It seems reasonable, 
therefore, in calculating the ultimate strength of riveted joints, 
to take as the strength of the material the minimum strength to 
which it has to be tested. Thus, in a ship built for the British 
Admiralty, we can use 26 tons as the strength per square inch 
of section, and in a ship built under Lloyd's rules, we can use 
28 tons per square inch of section. 

The following two examples will illustrate the methods 
adopted in calculating the strength of butt fastenings : 

i. A steel stringer plate is 48 inches broad and f 5 inch thick. Sketch 
the fastenings in a beam and at a butt, and show by calculations that the 
butt connection is a good one. 

(S. and A. xam., 1897.) 

For a fg-inch plate we shall require f -inch rivets, and setting these out 



240 Theoretical Naval Architecture. 

at the beam, we require 9 rivets, as shown in Fig. 84. The effective breadth 
of the plate through this line of holes is therefore 

48 9(i) = 41* inches 
and the strength is 

41^ X ^ X 26 = 470 tons 

and this is the standard strength that we have to aim at in designing the 
butt strap. 

(1) As regards the number of rivets. The shearing strength of a $-inch 
rivet being 1 1 '5 tons, the number of rivets necessary to equal the standard 
strength of 470 tons is 

470 

-^-^- = 40*8, say 41 rivets 

If we set out the rivets in the strap as shown in Fig. 84, leaving the 
alternate rivets out in the line AA, it will be found that exactly 41 rivets 
are obtained, with a four-diameter pitch. So that, as regards the number 
of rivets, the butt connection is a good one. 

(2) The strength of the plate in the line A A is the same as at the beam, 
the same number of rivet-holes being punched in each case. 

(3) If the strap is / s inch thick, the strength of the strap in the line BB 
is given by 

(48 i6(5)} x / fi X 26 = 410 tons 

This is not sufficient, and the strap must be thickened up. If made \ inch 
thick, the strength is 

(48 - i6(fl} X i X 26 = 468 
which is very nearly equal to the standard strength of 4/0 tons. 

(4) The shear of the 9 rivets in the line A A is 103-5 tons so tnat tn e 
strength of the plate through the line of holes CC and the shear of the 
rivets in the line AA are 

410 + 103-5 = 5 T 3'5 tons 

(5) Similarly, the strength of the strap through the line CC and the 
shear of the rivets in the line BB are 

468 + 184 = 652 tons 

The ultimate strengths of the butt connection in the five different ways it 
might break are therefore 47 1, 470, 468, 513^, 652 tons respectively, and 
thus the standard strength of 470 tons is maintained for all practical 
purposes, and consequently ihe butt connection is a good one. 

2. If it were required to so join two plates as to make the strength at 
the butt as nearly as possible equal to that of the unpierced plates, what 
kind of butt strap would you adopt ? 

Supposing the plates to be of mild steel 36 inches wide and inch thick, 
give the diameter, disposition, and pitch of rivets necessary in the strap. 

(S. and A. Exam., 1895.) 

The first part of this question has been already dealt with on p. 237. 
To lessen the number of rivets, it is best to use a double butt strap, as 
Fig. 85, so as to get a double shear of the rivets. Each of the butt straps 
should be slightly thicker than the half-thickness of the plate, say T 5 S inch. 

The standard strength to work up to is that of the plate through the 
single rivet-hole at the corner of the strap. |-inch rivets being used, the 
standard strength is 

(36 - I) X X 26 = 457 tons 



Calculation of Weights, etc. 241 

The single shear of a |-inch rivet is 15! tons, and the double shear may be 
taken as 

15-25 x i'8 = 27^ tons 

and consequently the least number of rivets required each side of the 
butt is 

452 J6-6, say 17 rivets 

The strength of the plate along the slanting row of holes furthest from 
the butt must be looked into. The rivets here are made with a water-tight 
pitch, say from 4 to 4^ diameters. If we set out the holes for a strap 2 feet 
wide, it will be found that the strength is below the standard. A strap 
3 feet wide will, however, give a strength through this line of about 465 
tons, which is very near the required 457 tons. There are 13 rivets along 
the edge of the strap, and the inside may be filled in as shown, giving a 
total number, of rivets, each side of the butt, of 19. 

For the strength of an assemblage of plating like the outer 
bottom, we must take the strakes as assisting one another. If 
two passing strakes are assumed, then we can take a butt with 
a through strake each side, and see how the strengths by 
various methods of fracture compare with the standard strength 
at a frame. 

For the strength of plating at a watertight bulkhead, the 
bulkhead liner is associated with the outside strake and one- 
half the adjacent inside strakes, and the strength should be 
brought up to that at an ordinary frame. 

Professor Hovgaard, in " Structural Design of Warships," 
deals very exhaustively with the above. In particular he 
allows for the reduction of area caused by countersinking and 
the slightly greater diameter of hole in the plate as compared 
with the nominal diameter of the rivet. 

Strength of Davits. The size of davits for merchant 
vessels are usually fixed by the rules of a Registration Society. 

The following is the rule adopted by Lloyd's Register, viz. : 

For boats and davits of ordinary proportions the diameter 
in inches is one-fifth of the length of the boat in feet. 

Where the height and spread of davits or dimensions of 
boats are not of ordinary proportions the diameter of davit 
in inches is found by the formula 



= VLXBXD/H X 

V 40 \3 / 



242 Theoretical Naval Architecture. 

Where L.B.D. are the dimensions of the boat, H is the 
height and R the outreach from the point of support in feet. 

The rule of the British Corporation is of the same form 
but slightly different, viz. : 



It is usual in H.M. service to test a davit to twice its 
working load, and this test load is used to calculate the dimen- 
sions. If W be the load in tons, r the outreach in inches, then 
the bending moment is W x r inch tons, and we apply the 
formula 

= y to find the diameter d, (y = J . d). 



/w7> 

'- 



Example. A boat weighing 2 tons is carried in davits with an out- 
reach of 6 ft. 6 in. Determine the diameter of davit, allowing a stress on 
the material of 5 tons per square inch. 

The moment WXr=2X78 = i56 inch tons, and the diameter is 
given by 



Example. A boat weighing I ton is carried in davits with an out- 
reach of 6 ft. 6 in. Determine the diameter of davit, allowing a stress on 
the material of 5 tons per square inch. 

Moment = 78 inch tons 



This davit was made sJ in. diameter, and 3$ in. at head and heel, the 
bow of davit being flattened out to an oval shape 5^ in. x 4f in. 

Example. A davit with outreach of 7 ft. 6 in. is tested to 3 tons. 
Find the maximum compressive and tensile tresses, the diameter of davit 
being 7 in. 

The max. BM is 

3 x 90 = 270 inch tons 

I of cross-section = V s . ^ . 49 = ir . 49 * 49 

4 64 

y = 3'S 

. * M 270 X 3-5 X 64 

'^ = T'^ = ^9x49 = 8topg **" 



Calculation cf Weights, etc. 243 

There is an additional compressive force due to the weight, viz. 

3 -i - = = *o8 tons sq. in. The tensile stress will be dimin- 

4 49-' 7r 
ished by this amount. 

/. Compressive stress = 8'o8 tons sq. in. 
Tensile stress = 7*92 tons sq. in. 

Davit Diagram. The following method of drawing once 
for all a davit diagram has beea brought to the Author's notice 
by J. J. King-Salter, Esq., R.C.N.C. Its use is very simple 
and obviates the necessity of calculating davit and similar 
diameters. 

If units are taken in Ibs. and inches, say a weight of w Ibs. 
and an overhang of r inches, then if a working stress is 
assumed of 4*5 tons per square inch, the diameter of a davit 
can be expressed in the simple form 



, // X r t . 

* m \* ' ...... (i) 

V 1000 

This is seen to depend on the product w X r. It can readily 
be shown that for a right-angled triangle, where a is the per- 
pendicular from the right angle t< ) the opposite side and b and 
c are the divisions of that side by the perpendicular, then 

a? = b X c\ or = *] b .c 

In Fig. $5A, above the base t the side is set up a scale of 
overhang in inches r> and below a scale of weight in Ibs. w. 
Along the base is set off a scale of *Jw .r. Thus for w = 
10,000 Ibs. and r - 100, */w.r=iooo. This point must 
subtend a right angle to the values of w and r taken, and this 
will determine the scale to use along the base. 

Now at various points along the base set up the corre- 
sponding value of d from the formula (i) above. Thus where 
V w . r 1000, w . r = 1,000,000 and d = 10 " ; d = 5" at an 
abscissa of 354, and so on. Through the spots thus obtained 
a curve of diameters is drawn as shown. 

The method of use is to employ a set square with the sides 
forming the right angle passing through the values given by 
the problem for w and r, with the right angle on the base line. 



Theoretical Naval Architecture. 



An ordinate from this point to the curve will give the diameter 
required. Thus for w = 4000 Ibs., r = 70", d is found to be 
6*6 in. If diameter is 8 in. say, and overhang is 80 in., the 

load is 6400 Ibs., and 
so on. 

The diagram can be 
drawn out on a large 
scale and mounted for 
general drawing office 
use. 

Pillars. Gordon's 
Formula. The for- 
mula usually employed 
to determine the strength 
of pillars is that known 
as Gordon's formula, as 
follows : 

W is the crippling 
load, A the cross sec- 
tion, f the stress given 

10,000 * in table, n is obtained 

from I = n . A . > 2 , where 
h is the least breadth, 




5.000 



c is a coefficient given in the table. 



W 



I -\ 75 

c.n.tf 



n = ^ for a rectangular section and yg for circular section ; for 
a circular hollow section n = |. 

Units are taken as tons and inches. 



Material 


/ 


c 




tons sq. in. 


Ends free. 


One end fixed. 


Both ends fixed. 


Wrought iron . 
Mild steel . . 


16 

3 


9,000 

9,OOO 


18,000 
18,000 


36,000 
36,000 


Cast iron . . 


35 


1,600 


3,200 


6,400 


Dry timber . . 


3 


750 


1,500 


3,ooo 



Calculation of Weights, etc. 



245 



This formula is empirical. For a discussion regarding its 
use the student is referred to such works as Lineham's 
" Mechanical Engineering." 

Example. A cargo derrick for a vessel is constructed of steel plating 
5 ^5 in. thick and two T bars 5 in, X 3 in. X 5 8 in. The jib is 40 ft. long, and 
the topping lift is led to a point on a mast 32 ft. above heel of derrick. The 



/S 




FIG. 858. 



maximum load to be lifted being 16 tons ; calculate the approximate 
diameter of the derrick if the maximum stress on the material is not to 
exceed 4 tons per sq. in. (neglect effect of T bars). 

(Honours B. of E. 1909.) 

By setting out the triangle of forces CDE at the head of derrick the 
thrust on derrick is found to be 20 tons, DE being parallel to CB (Fig. 



In the above formula f 4, c 
the unknown diameter. 



9000, n J, A = - .d % d being 



246 Theoretical Naval Architecture. 

'----* ~ 

"!*.</ ,.+ 



9000 x | 



or o-25^ 3 <f>^,&\$ 

By trial d is found to be 1 1 inches nearly. 

Example. A wooden derrick 34 ft. long when tested to twice the 
working load is found to be subject to a thrust of 4^ tons. Determine the 
diameter, allowing a factor of safety of 6 when being tested. 




26 
TONS 



FIG. 850. 



In the above formula, W = 25-5 tons, /= 3, c - 750, / = 34 X 12, 
n T l s , d diameter. 



Tr.a' (34 X 12)' 

4 750 X T ' 8 X a 



= 0-092^ 



or cro92</ 4 - d? = 3550 
By trial d is found to be about 14^ in. 



Calcidation of Weights, etc. 247 

Example. A boat hoisting derrick 60 ft. long, estimated weight 6 tons, 
is arranged as shown on figure herewith, the purchase being single through 
sheaves A and D. The topping lift has the fixed part at E, and passes 
through sheaves B and C. Determine the forces on blocks and ropes and 
the thrust on derrick, when holding the test load of 26 tons. Determine 
the diameter of the derrick if formed of in. steel plating, the T bars 
forming edge strips being neglected, and a factor of safety of 5 being 
assumed. (Fig. 850.) 

oa is set down = 29 tons, i.e. 26 tons plus half weight of derrick, and 
ab is drawn parallel to the purchase AD, and equal to 26 tons. Then ob 
is the resultant force at the head of derrick due to the forces on the purchase 
and the weight of derrick, be is then drawn parallel to the topping lift. 
Then obc is a triangle of forces, giving for the force on topping lift 
be 35 tons and the thrust on derrick oc = 59 tons. The force in the link 
AB is therefore 35 tons, and the two parts of topping lift have each a 
force of 17*5 tons. The force on the block C is found by drawing the 
triangle of forces deg, de = eg=\T$ tons from which force on block 
is dg 29 tons. Similarly the force on block D is found to be 
38 tons. 

The length of derrick from pin of sheave to the trunnion is 56*5 feet. 
In Gordon's formula we have therefore 

W = 59 X5 = 295 tons,/= 30, c 9000, n = J, d y the diameter, is the 
unknown, A = . TT . </, / = 56*5 X 12. 

We have therefore 

30 X j x y x d 
295 ~ ! , (56-5 x 12)* 

9000 X 5 X d* 

410 
or I + -^ - = O'l2d 

or o'izd 3 d? 410 
from which d is found by trial to be i8 ins. nearly. 

It may be noted that the above derrick was actually made 
20 in. diameter, of plating, 14 Ibs. per square foot, which allows 
for the loss due to the rivets in butt strap. 

Fig. 850 illustrates the case where electric winches are 
employed for both topping lift and purchase, and the 
greater speed of these winches renders more turns of rope 
necessary. 

The test load is 32 tons, being 'twice the weight of the 
boat. This with the weight of the block A gives 32-4 tons, 
which is taken by the three ropes supporting A, giving 10-8 
tons to each. To find the force at the topping lift and on 
the derrick, we draw the diagram of forces, shown on top 
of figure, ab = 36 tons, i.e. 32 + 3-6 (half weight of der- 
rick) + 0-4 (weight of block A) ; be io'8 is drawn parallel 



248 



Theoretical Naval Architecture. 



to the purchase, so that ac is the force at the head of the 
derrick ; cd is drawn parallel to the topping lift, and ad 
parallel to the derrick. Then cd is force on topping lift = 59 
tons, and ad the thrust on derrick = 60 tons. The topping 
lift is in four parts, giving 15 tons to each, and the block C has 
59 tons. The block D has three parts, viz. 45 tons. The block 
E has 15 tons along each of the ropes, and as shown by the 







72 TONS 




FIG. 850. 

diagram, sustains a force of 26 tons. The block F has 10*8 
tons along each of the ropes, and, as shown by the diagram, 
sustains a force of 18 tons. The strength at the heel of the 
derrick can be allowed for knowing the thrust to be 60 tons. 
The test loads of the various parts can now be allowed for, 
being usually about twice that due to the 32-ton test load 
applied. 



Calculation of Weights, etc. 



249 



Fig 85 E shows an ordinary form of derrick for a cargo 
vessel, the derrick being supported by a stump mast. The 
purchase is taken round two single blocks A and B, and thence 



DOUBLE 
D 




FIG. 858. 

to the winch. The topping lift has a single block at C, a 
double block at D, and a single block at E, and thence to the 
winch. Taking a load of 5 tons the triangle of forces abd is 
drawn ab ac 5, and the resultant force due to the load is 
ad 9 tons, which is the force on the block A. de is drawn 
parallel to the topping lift. Then ae = 14 tons is the thrust 



250 



Theoretical Naval Architecture. 



on the derrick, and 67 tons is the force on block C. This 
divided into 3 gives 2*2 tons in each portion of the topping 
lift. The force on block D is obtained by drawing the 
triangle of forces fgh. Similarly the forces on the blocks B 
and E are obtained. The forces thus obtained give a basis 
for estimating the strength of all the parts, including the 
derrick and the stump mast. 

Shaft Brackets. (By A. W. Johns, Esq., R.C.N.C.). 
The length and diameter of the drum or barrel of a shaft 
bracket are determined by the requirements of the engineer. 
The inside diameter is arranged to take the shaft and its 




bearings and bushes. The outside diameter is usually from 
3 to 6 inches greater than the inside diameter, depending on 
the size of shaft The inside surface is generally gulleted to 
a depth of from i to ij inches. The length of barrel is 
governed by the length required by the engineer's bearings 
in the bracket (see Fig. 85 F). 

The length of the arms or struts must necessarily depenc 
on the position of the axis of the shaft at the bracket and the 
shape of the ship in the vicinity. The section of the arms is 
usually pear-shaped (see Figs. 850 and 108 for examples) 
with the blunt end forward. The dimensions of the section 
must be governed by the straining action to which the 
bracket is subjected. Formerly these dimensions appear 



Calculations of Weights* etc. 251 

to have been determined in a rough-and-ready way from the 
experience of the designer responsible. Knowing the dimen- 
sions in previous cases which on service had proved sufficiently 




FIG. 850. 

strong, he would vary these dimensions in a new ship accord- 
ing to the variation of the horse-power, or perhaps the size and 
overhang of the tail shaft. Consequently it will be found that 
ships of about the same size, horse-power and revolutions pro- 
duced under different designers, have entirely different dimen- 
sions (and weights) of shaft brackets. 

At first sight a suitable basis of comparison for such dimen- 
sions appears difficult to obtain, but investigation proves that 
the matter is a comparatively simple one, as is seen by what 
follows : 

With the centre of gravity of the revolving parts, viz. shaft 
and propeller, in the axis of rotation, the straining actions 
which may operate on a bracket are as follows, viz. : 

1. Forces due to the weight of the propeller, shaft 
and bracket. These are equivalent to a downward force 
on the bracket, and a bending moment on it, equal to the 
difference in the moments of weight on the forward and 
after sides. Both the force and the bending moment may 
be increased appreciably by the accelerative effect during 
pitching. 

Thus in a given ship 500 feet long pitching in a " single " period of 
3 seconds, the maximum acceleration is 250 X , . X 0, where is angle of 

pitching (see Chap. IX. on Rolling). 

If 9 = 4, say J 4 in circular measure, acceleration = 20 in foot-second 
units. This added to the acceleration due to gravity gives 52^2 or a 
virtual weight of I '6 times the actual, and all the forces are increased in 
this ratio. 

2. Forces called into play when pitching or turning due to 
the gyroscopic action of the propeller and shaft. 



252 



Theoretical Naval Architecture. 



3. Forces caused by unequal pressure on the blades of the 
propeller when the ship is turning. Here, owing to the trans- 
verse motion of the stern, the forces on the blades above the 
horizontal will be different to those below. 

If, however, the centre of gravity of the revolving weights 
is not in the axis of rotation, there will be in addition to the 
above forces a centrifugal force operating which will tend to 
bend the shaft where it enters the strut, and will also tend to 
bend and twist the bracket. At high revolutions, which is the 
case in turbine machinery, heavy straining actions are set up 
if a propeller blade is broken or lost. 

The following are approximate values of the various forces 
and moments considered above, worked out for the case of 
a large cruiser. 





Force on bracket. 


Moment on bracket. 


(i) Due to weight of propeller, 
etc 


30 tons 


90 foot tons 


(la) Due to weight of propeller, 
etc., when pitching . . 
(2) Gyroscopic action . . . 
(3) Turning at full speed with 
full rudder angle . . . 
(4) Centrifugal action due to 
the loss of a propeller 
blade at full revolutions 


48 

12 
100 


M4 
5 

80 

5 ' 



It will be seen that the last case produces by far the 
heaviest straining effect. In addition to straining the bracket, 
however, the shaft also will be strained, the maximum stress 
occurring at the section immediately at the after end of the 
barrel of the bracket. For good design the bracket should 
be stronger than the shaft, for then the shaft would break at 
the after end of the boss of shaft bracket, whereas if the 
bracket were weaker than the shaft the former would first 
break and the shaft losing its after support would then bend 
or break with perhaps disastrous effect on the ship. A basis 
of calculation is therefore obtained by considering the strength 
of the shaft and making the shaft bracket somewhat stronger. 

If T is the force at the propeller (Fig. 85 F), producing 



Calculation of Weights, etc. 253 

a bending moment T . a on the shaft which will just bring the 
material of the shaft to its full working strength, this force T 
must be employed in determining the dimensions of the arms 
of the bracket and also in determining the number, size 
and spacing of the rivets connecting the bracket to the hull. 
T acting at propeller is equivalent to 

1. A parallel force T acting directly on the bracket, and 

2. A moment k . T . m on the bracket, where k has an 
average value of about 0-65. 

If T is caused by centrifugal action, then as the shaft 
revolves it is always being bent in the same way, but the 
bracket being fixed the force and moment on it are constantly 
altering in direction. 1 Bending alone occurs when the line of 
action of T lies in a plane passing through the axis of shaft 
and bisecting the angle between the arms. Twisting occurs 
when the line of action of T is perpendicular to that plane. 
For other directions of the line of action of T combined bend- 
ing and twisting occur. 

Generally bending alone produces the greatest stress on 
the shaft arms, and this produces a stress given by 
.T. m. y.cos 

p ~- ~ir 

where I is the moment of inertia of a right section of the arm 
about an axis through the geometrical centre and 
perpendicular to the longer dimension of the section. 
y is the distance of the most strained layer from this axis ; 
and 6 is half the angle between the arms. 
For ordinary pear-shaped sections,^ = 0*55 R and I = gV-'R 8 . r, 
where R and r are the longer and shorter dimensions of the 
section of the arm. 

Taking, say, 6 tons as the maximum working strength of 
the shaft, the force T necessary to strain the shaft to this limit 
can readily be found when a the overhang and D and d the 
external and internal diameters of the shaft are known. Sup- 
posing the shaft bracket is of cast steel and taking 4^ tons as 
the working strength (5 tons is really allowed, but \ ton is 

1 The loss of a propeller blade is soon evident, for the ship will vibrate 
violently if the revolutions of the engine approach the full number. 



254 



Theoretical Naval Architecture. 



allowed for the force T acting directly on the bracket), the 
following relation is obtained 

R 2 .r=o-63X ^ x^Xcosfl . . (i) 

All dimensions being in inches. 

Usually 6 = about 45 and we then have 



D 4 - 



m 



R 2 . r = 0-44 X ^ X - 



If, however, 9 is small, we have approximately- 

D 4 -d* m 
R 2 . r = 0-63 X 



D 



<*) 



(3) 




FIG. 



As stated above, the stress produced on the bracket by 
bending is usually greater than that 
produced by twisting, but in the case 
where the angle between the arms is 
small the stress due to twisting should 
also be investigated. This can be 
done as follows : 

Taking, as in the figure 85 H, was 
the distance between the centres of 
the arms, A the area of each arm, 
q the stress in the arm, the moment 
resisting twisting is given by q . A . n. 
This must equal k . T . m, and hence the stress due to twisting 
becomes 

k .T. m . A 

q= A n (A = 

q is a shearing stress and the equation 
shows that if A is constant q increases as 
n decreases. Close to the barrel q is a 
maximum, and is a minimum where the 
arms enter the hull. For q to be constant 
A should vary inversely as n. Usually, 
however, A is kept constant and the arms 
are run tangential, as in Fig. 85;, instead of radial to the 
barrel. This has the effect of increasing n near the barrel and 
diminishing q. 

Equations (i), (2), and (3) above can be used to determine 




FIG. 85;. 



Calculation of Weights, etc. 



255 



the value of R 7 r, where the bracket is of cast steel. If the 
bracket is of other material the working strength of the latter 
must be substituted for the 4^ tons used above. 

It will be noticed that economy of material is obtained by 
making the ratio R -r r as large as possible, for since the 
square of R enters into the relation it has far more influence 
than r, which appears in the first power only. Thus if R a . r 
= 8000 and R = $r, then R = 29 in. and r = Q in. and A 
= 212. Whereas if R = 6r, R = 36 in., and r = 6 in., and 
A = 162 or a saving in weight of about 25 per cent. There 
is also an appreciable reduction in resistance. Taylor's ex- 
periments with shaft brackets show that resistance in Ibs. per 
foot length of shaft bracket arm is given by 

F = -^-(A + 4o)V a 

IOOO V 

where V = speed in knots 

A = area of section in square inches, for values between 
40 and 175 sq. inches. 

c = a constant depending on the ratio . 
VALUES OF c. 



Ratio Jl 
r 


3 


4 


5 


6 


7 


8 


9 


10 


ii 


12 
























Value of c 


1-88 


1-32 


1-07 


0-94 


0-86 


0-80 


076 


074 


072 


071 



Further than this there is little doubt that with bracket 
arms whose ratio R -r- r is small, at high speeds a large amount 
of dead water trails behind the arms reaching to the propeller 
disc and causing vibration and loss of propeller efficiency as 
the blades of the propeller enter and leave the dead water. 
By increasing the ratio R -4- r these effects are diminished. 
Fig. 85 E gives the section of the arm of shaft bracket of a 
recent ship of large power and great speed. 

Finally it is interesting to compare the dimensions given 
by formula (2) above with those adopted in practice in parti- 
cular cases as the result of experience. The ratio R -f- r has 



254 



Theoretical Naval Architecture. 



allowed for the force T acting directly on the bracket), the 
following relation is obtained 



= 0-63 X 



D 4 - 
D 



m 
x -Xcos0 



(0 



All dimensions being in inches. 

Usually 6 = about 45 and we then have 

D 4 d* m 
r = 0-44 : 



R 2 



1) 



If, however, 9 is small, we have approximately 

D* -d* m 
13*. r= 0-63 X ^ X - 



(3) 



D a 

As stated above, the stress produced on the bracket by 
bending is usually greater than that 
produced by twisting, but in the case 
where the angle between the arms is 
small the stress due to twisting should 
also be investigated. This can be 
done as follows : 

Taking, as in the figure 85 H, was 
the distance between the centres of 
the arms, A the area of each arm, 
q the stress in the arm, the moment 
resisting twisting is given by q . A . n. 
This must equal k . T . m, and hence the stress due to twisting 
becomes 

.T. m 




FIG. 




q is a shearing stress and the equation 
shows that if A is constant q increases as 
n decreases. Close to the barrel q is a 
maximum, and is a minimum where the 
arms enter the hull. For q to be constant 
A should vary inversely as ;/. Usually, 
however, A is kept constant and the arms 
are run tangential, as in Fig. 85;, instead of radial to the 
barrel. This has the effect of increasing n near the barrel and 
diminishing q. 

Equations (i), (2), and (3) above can be used to determine 



FIG. 8sj. 



Calculation of Weights, etc. 



255 



the value of R 7 r, where the bracket is of cast steel. If the 
bracket is of other material the working strength of the latter 
must be substituted for the 4^ tons used above. 

It will be noticed that economy of material is obtained by 
making the ratio R -4- r as large as possible, for since the 
square of R enters into the relation it has far more influence 
than r, which appears in the first power only. Thus if R a . r 
8000 and R = 3r, then R = 29 in. and r = Q|- in. and A 
= 212. Whereas if R = 6r, R = 36 in., and r = 6 in., and 
A = 162 or a saving in weight of about 25 per cent. There 
is also an appreciable reduction in resistance. Taylor's ex- 
periments with shaft brackets show that resistance in Ibs. per 
foot length of shaft bracket arm is given by 

F = (A + 40) V a 

IOOO 

where V = speed in knots 

A = area of section in square inches, for values between 

40 and 175 sq. inches. 

-n 

c a constant depending on the ratio . 
VALUES OF c. 



Ratio IL 
r 


3 


4 


5 


6 


7 


8 


9 


10 


ii 


12 


Value of c 


1-88 


1-32 


1-07 


0-94 


0-86 


0-80 


076 


074 


072 


071 



Further than this there is little doubt that with bracket 
arms whose ratio R -4- r is small, at high speeds a large amount 
of dead water trails behind the arms reaching to the propeller 
disc and causing vibration and loss of propeller efficiency as 
the blades of the propeller enter and leave the dead water. 
By increasing the ratio R -f- r these effects are diminished. 
Fig. 85 E gives the section of the arm of shaft bracket of a 
recent ship of large power and great speed. 

Finally it is interesting to compare the dimensions given 
by formula (2) above with those adopted in practice in parti- 
cular cases as the result of experience. The ratio R -f- r has 



2 5 6 



Theoretical Naval Architecture. 



been kept the same in the calculation as actually adopted. 
All dimensions are in inches. 













Actually fitted. 


Calculated. 












R 


r 


R 


r 


Cruiser . 


7f 





37 


12 


14 


4 


13-3 


ri 






9 




21 


16 


6 


16-8 


6-3 




16 


10 
10 

7 


5 
62 

78 


27 
27 

33 


20 

11 


6 
6 
8 


2i'5 

24 
30-1 


! 

8-6 




21 


ii 


76 


30 


32 


10 


31-2 


9-8 


Battleship 
Destroyer 


20 

84 


4* 


72 
36 


18 


27 
13 


2 


30 


2'2 



The relations in d), (2), and (3) given above apply strictly 
to the usual pear-shaped section of arm, but the method indi- 
cated can be applied to any particular case and the dimensions 
calculated. 

EXAMPLES TO CHAPTER VI. 

1. The area of the outer bottom plating of a ship, over which the 
plating is worked 25 Ibs. per square foot, is 23,904 square feet, lapped 
edges and butt straps, both double-riveted. Estimate the difference in 
weight due to working the plating with average-sized plates 20' X 4^', or 
with the average size 12' x 3'. 

Ans. About 20 tons. 

2. Steel angle bars 3^" X 3" are specified to be 8 Ibs. per lineal foot 
instead of ^ inch thick. Determine the saving of weight per 100 lineal 
feet. 

Ans. 52 Ibs. 

3. Determine the weight per lineal foot of a steel T~bar 5" x 4" X J". 

Ans. 14-45 lbs 

4. For a given purpose, angle bars of iron 5" X 3" X T 8 8 " or of steel 
5" X 3" X 2 V' can be used. Find the saving of weight per 100 feet if steel 
is adopted. 

Ans. 95 Ibs. 

5. A mast 96 feet in length, if made of iron, is at its greatest diameter, 
viz. 32 inches, ^ inch thick, and has three angle stiffeners, 5" X 3" x ^". 
For the same diameter, if made of steel, the thickness is 55 inch, with three 
angle stiffeners 5" X 3" X 5 9 ". Estimate the difference in weight. 

Ans. About I ton. 

6. At a given section of a ship the following is the form : The lengths of 
ordinates 3 feet apart are 19*6, 18-85, r 7'8, 16*4, 14-5, 1 1 '8, 7-35, and 
I'D feet respectively. Estimate the vertical position of the centre of 
gravity of the curve forming the section, supposing it is required to find the 
vertical position of the centre of gravity of the bottom plating of uniform 
thickness. 

Ans. About 12\ feet from the top. 



Calculation of Weights, etc. 257 

7. The half-girths of the inner bottom of a vessel at intervals of 51 feet 
are 26*6, 29*8, 32-0, 32 '8, and 31*2 feet respectively, and the centres of 
gravity of these half-girths are i8'6, 20*6, 21 '2, 20*0, 17-4 feet respectively 
below the L.W.L. Determine the area of the inner bottom and the 
position of its centre of gravity both longitudinally and vertically. If the 
plating is of 15 Ibs. to the square foot, what would be the weight, allowing 
14^ per cent, for butts, laps, and rivet-heads. 

Ans. 12,655 square feet ; 105 feet from mer end, 2O feet below the 
L.W.L. ; 97 tons. 

8. The whole ordinates of the boundary of a ship's deck are "S'5, 24, 
29, 32, 33'5, 33'5, 33'5> 32, 3 2 7, and 6-5 feet respectively, and the 
common interval between them is 21 feet. 

The deck, with the exception of 350 square feet, is covered with \ inch 
steel plating worked flush jointed, with single riveted edges and butts. 
Find the weight of the plating, including straps and fastenings. 

Ans. 45 tons. 

9. A teak deck, 2^ inches thick, is supported on beams spaced 4 feet 
apart, and weighing 15 pounds per foot run. Calculate the weight of a 
middle-line portion of this deck (including fastenings and beams) 24 feet 
long and 10 feet wide. Ans. I '65 tons nearly. 

10. Taking the net weight of outer bottom plating of a vessel as 1000 
tons, estimate the saving of weight if the average size of plates is 20 feet 
by 5 feet as against 18 feet by 3! feet. (Butt-straps double riveted, lapped 
edges double riveted, -inch rivets.) Ans. 43 tons about. 

n. A longitudinal W.T. bulkhead is bounded at its upper edge by a 
level deck (having 9-inch beams, 4 feet apart) and at its lower edge by the 
inner bottom. The depths of the bulkhead at ordinates 61 feet apart are, 
commencing from forward, g'o, 16*7, 19*3, I5'4> 9'5 feet respectively. 
The plating of the bulkhend is 15 Ibs. per square foot worked vertically, 
single riveted, and the stiffening consists of Z bars of 12 Ibs. per foot 
spaced 4 feet apart with intermediate angles of 7 Ibs. per foot. There is 
a single boundary bar of 8'5 Ibs. per foot. 

Calculate (i) the weight of the bulkhead. 

(2) the distance of C.G. from forward end. 

(3) the distance of C.G. below the deck. 

Ans. (i) 39 tons ; (2) 120-5 feet ; (3) 8 feet - 

12. The half ordinates of upper deck of a ship 360 feet long are (i) o ; 
(2) 9'4; (3) l6 ' 2 ; (5) 24-4; (7) 28-8; (9) 31-2; (ii) 32-4; (13) 32-2; 
(15) 31-5 ; (17) 29-6 ; (19) 24 6 ; (20) 20'i ; (21) 13-8. Over the midship 
portion (7) to (15) the beams are 24^ Ibs. per foot, 4 feet apart, and the 
plating is 20 Ibs. per square foot, with single-riveted edge-strips and 
double-riveted butt-straps. At the ends the beams are 24^ Ibs., 3 feet 
apart, completely covered with plating 10 Ibs. per square foot, lapped and 
single-riveted. The boundary bar is 3 inches by 3! inches of 8 Ibs. per 
foot, and the deck is completely planked with 3-inch teak. Find total 
weight, neglecting hatches, etc. 

Ans. 325 tons about. 



CHAPTER VII. 

STRAINS EXPERIENCED BY SHIPS CURVES OF LOADS, 
SHEARING FORCE, AND BENDING MOMENT- 
EQUIVALENT GIRDER "SMITH" CORRECTION 
TROCHOIDAL WAVE. 

Strains experienced by Ships. The strains to which ships 
are subjected may be divided into two classes, viz. 

1. Structural strains, i.e. strains which affect the structure 
of the ship considered as a whole. 

2. Local strains, i.e. strains which affect particular portions 
of the ship. 

1. Structural Strains. These may be classified as 
follows : 

(a) Strains tending to cause the ship to bend in a fore-and- 
aft direction. 

(b) Strains tending to change the transverse form of the 
ship. 

(c) Strains due to the propulsion of the vessel, either by 
steam or sails. 

2. Local Strains. These may be classified as follows : 

(a) Panting strains. 

(b) Strains due to heavy local weights, as masts, engines, 
armour, guns, etc. 

(c) Strains caused by the thrust of the propellers. 

(d) Strains caused by the attachment of rigging. 
((?) Strains due to grounding. 

We will now deal with some of these various strains to 
which a ship may be subjected in a little more detail. 

Longitudinal Bending Strains. A ship may be regarded as 
a large beam or girder, subject to bending in a fore-and-aft 
direction. The support of the buoyancy and the distribution 
of weight vary considerably along the length of a ship, even 



Strains experienced by Ships, etc. 



259 



when floating in still water. Take a ship and imagine she is 
cut by a number of transverse sections, as in Fig. 86. Each 
of the portions has its weight, and each has an upward support 
of buoyancy. But in some of the portions the weight exceeds 
the buoyancy, and in others the buoyancy exceeds the weight. 
The total buoyancy of all the sections must, of course, equal the 
total weight. Now imagine that there is a water-tight bulkhead 
at each end of each of these portions, and the ship is actually 
cut at these sections. Then the end portions (i) and (5) have 
considerable weight but small displacement, and consequently 
they would sink deeper in the water if left to themselves. 1 In 



i 




FUG. 86. 

the portions (2) and (4), on the other hand, the buoyancy might 
exceed the weight (suppose these are the fore-and-aft holds, and 
the ship is light), and if left to themselves they would rise. The 
midship portion (3) has a large amount of buoyancy, but also 
a large weight of engines and boilers, and this portion might 
very well have to sink a small amount if left to itself. In any 
actual ship, of course, it is a matter of calculation to find how 
the weight and buoyancy vary throughout the length. This 
case is somewhat analogous to the case of a beam supported 
and loaded as shown in Fig. 87. At each point along the 
beam there is a tendency to bend, caused by the way the 
beam is loaded and supported, and the beam must be made 

1 Strictly speaking, each portion would change trim if left to itself, but 
we suppose that the various portions are attached, but free to move in a 
vertical direction. 



260 



Theoretical Naval Architecture. 



sufficiently strong to withstand this bending tendency. In the 
same way, the ship must be constructed in such a manner as 
to resist effectually the bending strains that are brought to 
bear upon the structure. 

When a vessel passes out of still water and encounters 



FIG. 87. 

waves at sea, the strains to which she is subjected must differ 
very much from those we have been considering above. 
Suppose the ship to be end on to a series of waves having 
lengths from crest to crest or from trough to trough equal to 
the length of the ship. We will take the two extremes. 

(1) The ship is supposed to have the crest of the wave 
amidships. 

(2) The ship is supposed to have the trough of the wave 
amidships. 



HOGGING 





ACROSS WAVE TROUGH 

FIGS. 88, 89. 



(i) This is indicated in Fig. 88. At this instant there 
is an excess of weight at the ends, and an excess of 
buoyancy amidships. The ship may be roughly compared 
to a beam supported at the middle, with weights at the end. 



Strains experienced by Ships, etc. 



261 



as in Fig. 90. The consequence is that there is a tendency 
for the ends to droop relatively to the middle. This is termed 



(2) This is indicated in Fig. 89. At this instant there 
is an excess of weight amidships, and an excess of buoy- 
ancy at the ends, and the ship may be roughly compared 
to a beam supported at the ends and loaded in the middle, 
as Fig. 91. The consequence is, there is a tendency for the 
middle to droop relatively to the ends. This is termed 
sagging. 



FIG. oo. 

We have seen above how the ship may be compared to a 
beam, and in order to understand how the material should be 
disposed in order best to withstand the bending strains, we 
will consider briefly some points in connection with ordinary 
beams. 



FIG. 91. 

Take a beam supported at the ends and loaded at the 
middle. It will bend as shown exaggerated in Fig. 92. The 
resistance the beam will offer to bending will depend on 
the form of the section of the beam. Take a beam having 
a sectional area of 16 square inches. We can dispose the 
material in many different ways. Take the following : 



262 



Theoretical Naval Architecture. 



(a) 8 inches wide, 2 inches deep (a, Fig. 93). 
(<) 4 inches wide, 4 inches deep (, Fig. 93). 

(c) 2 inches wide, 8 inches deep (<r, Fig. 93). 

(d) 8 inches deep, with top and bottom flanges 5 inches 
wide and i inch thick (d, Fig. 93). 




FIG. 92. 

Then the resistances of these various sections to bending 
compare as follows : 

If (a) is taken as i, then (b) is 2, (c) is 4, and (d) 
is6f. 

We thus see that we can make the beam stronger to resist 
bending by disposing the material far away from the centre. 




FIG. 93. 

The beam (d) has 6f times the strength of (a) against bending, 
although it has precisely the same sectional area. A line 
drawn transversely through the centre of gravity of the section 
of a beam is termed the neutral axis. 

In the British Standard Sections it will be found that for 
Z bars and channel bars the flanges are distinctly thicker 
than the web. 

These principles apply equally to the case of a ship, and 
we thus see that to resist bending strains the material of the 



Strains experienced by Ships, etc. 



263 



structure 
axis. 1 



should be disposed far away from the neutral 



For hogging strains, the upper portions of the vessel are 
in tension and the lower portions are in compression. For 
sagging strains, the upper portions are in compression and the 
lower portions are in tension. Thus the portions of the struc- 
ture that are useful in resisting these hogging and sagging strains 
are the upper and main decks and stringers, sheer strake and 
plating below, plating at and below the bilge, both of the inner 
and outer bottom, keel, keelsons, and longitudinal framing. 

Strains tending to change the Transverse Form of the Ship. 
Strains of this character are set up in a ship rolling heavily. 
Take a square framework joined at the corners, and imagine 
it to be rapidly moved backwards and forwards as a ship does 
when she rolls. The framework will not break, but will distort, 
as shown in Fig. 94. There is a tendency to distort in a similar 




FIG. 



way in a ship rolling heavily, and the connections of the beams 
to the sides, and the transverse structure of the ship, must be 
made sufficiently strong to prevent any of this racking taking 
place. Transverse bulkheads are valuable in resisting the 
tendency to change the transverse form. 

A. ship, when docked, especially if she has on board heavy 
weights, as armour or coals, is subjected to severe strains 
tending to change the transverse form. If the ship is supported 



1 There are other strains, viz. shearing strains, which are of importance 
(see later). 



266 



Theoretical Naval Architecture. 



points along the beam, we shall be able to draw a line through all 
the spots as amb, which has a maximum ordinate at the centre 
o of J . W . a. This line will give the bending moment at any 
point along the beam. 

Or take the case of a beam supported at the ends and 
loaded uniformly with the weight w per foot run, the total weight 



w.o/ 




being, therefore, 2 . w . a, as Fig.96. The support at each end 
is w . a. The bending moment at any point K, distance x from 
the end, is M k = w . a . x J . w . oc*. When x a this bend- 
ing moment is therefore J . w . a 2 . If a number of spots be 
thus obtained throughout the length of the beam, we can draw 
a curve as amb, any ordinate of which will give the bending 
moment at that point of the beam. 




FIG. 97. 

Take now the case of a beam supported at one end and 
loaded uniformly. The load can be graphically represented by 
a rectangle oabc> ab = w (Fig. 97). At any point K, calculate 



Strains experienced by Ships, etc. 



267 



the area of the rectangle ob on one side of k, i.e. w . x, and set off 
as an ordinate kf = w .x. This is what is termed the " shearing 
force " at K, or the tendency the two consecutive sections of 
the beam at K have to slide over one another. Doing this 
all along the beam, we should obtain the line afg> the maximum 
ordinate of which og = w .1. 

The area of the figure akf= .w.x*, and the bending momen* 
at K also = % . w . x?. So that to determine the bending momem 
at any point, we find the area of the curve of shearing force up 
to that point. 1 In this way the curve of bending moment 
amm is constructed, having a maximum ordinate om of \ . wl 2 . 
This method of determining the bending moment at any point 
from the curve of shearing force is of no value in this particular 
case, but is of assistance when dealing with more complicated 
cases cf loading. 

Take, for example, a beam similar to the above, but loaded 
unevenly along its length, such that the intensity of the load 




FIG. 98. 



at any point is given by the ordinate of the curve //, which we 
may term a " curve of loads'' as Fig. 98. Take any point K 
and determine the area beneath the curve of loads from the 
point k to the end of the beam. This will give the shearing 
force at K. Doing this all along the beam, we can draw the 
curve of shearing force aff. The area under this curve between 

1 For the proof of this in any general case, see any standard work on 
*' Applied Mechanics," as Cotterill, chap. iii. 



268 



Theoretical Naval Architecture. 



k and the end of the beam gives the bending moment at K, 
and in this manner the curve of bending moment amm Can 
be obtained. 

Turning now to the case of a ship floating in still water. 
There will be a certain distribution of the weight and also of 
the buoyancy. The total weight must, of course, be equal to 
the total buoyancy, and also the fore-and-aft position of the 
centre of gravity of the weight must be in the same athwartship 
section as the centre of buoyancy. But although this is so, 
the distribution of the weight and buoyancy along the ship 
must vary from section to section. 




FIG. 99- 

Take the case of a vessel floating in still water in which 
the buoyancy exceeds the weight amidships, and the weight 
exceeds the buoyancy at the ends. Let BB in Fig. 99 be the 
" curve of buoyancy." The area under this curve will give 
the displacement of the vessel, and the fore-and-aft position of 
the centre of gravity of this area is the same as the fore-and-aft 
position of the centre of buoyancy. 

Also let WW be the " curve of weight." This curve is 
constructed by taking all the weights between two sections and 
setting up a mean ordinate to represent the total weight between 
the sections. This done throughout the length gives a number 
of spots through which a curve may be drawn as nearly as 



Strains experienced by Ships, etc. 269 

possible. This curve should be adjusted as necessary to fulfil 
the conditions stated above, viz. that the area under it shall 
equal the area of the curve of buoyancy, and the fore-and-aft 
position of the centre of gravity of the area under it shall be 
in the same section as that of the curve of buoyancy. 

The difference at any point between the ordinates of the 
two curves WW and BB will give the difference between the 
weight and the buoyancy at that point. Where the curves 
cross at A and B, the weight and buoyancy are equal, and the 
sections at these points are said to be " water-borne" 

Now set off ordinates all along, giving the intercept between 
the curves WW and BB. Set below the base-line where the 
weight exceeds the buoyancy, and above the base-line where 
the buoyancy exceeds the weight. In this way we obtain the 
curve LLL which is the " curve of loads" At the sections 
where the curve of loads crosses the base-line the ship is 
water-borne. We now obtain the "curve of shearing force" 
FFF from the curve of loads by finding the area under the 
curve of loads, as explained above. Also in a similar manner 
the curve of bending moment MM is obtained by finding 
the area under the curve of shearing force. The maximum 
ordinate of this curve will give the greatest bending moment 
the ship will be subjected to under the assumed conditions. 

In constructing curves of bending moment, moments 
tending to cause " hogging " are put above the base-line, and 
moments tending to cause " sagging " are put below the base- 
line. In the case in Fig. 99 the moments are " hogging " 
throughout the whole length of the vessel. 

The area of the curve of loads above the base-line being 
the same as the area below the base-line, it follows that the 
ordinate of the shearing force must come to zero at the end. 
Also the ordinate of the curve of bending moment must be zero 
at the end, and this constitutes a most effective check on the 
accuracy of the work. 

It is obvious, however, that the strains due to the bending 
moment in still water are not the worst that in practice will 
affect the longitudinal structure of the ship. The strains in still 
water are small in magnitude compared with the strains that 



2/0 



Theoretical Naval Architecture. 



may affect the ship at sea. For a ship at sea there are two 
extreme cases that can be assumed, viz. 

(1) The ship being supposed to be momentarily at rest on 
the crest of a wave of her own length, the height of the wave 
being taken some proportion of the length (Fig. 100). 

(2) The ship being supposed to be momentarily at rest across 
the trough of a wave similar to that assumed in (i) (Fig. 101). 

It is usual to take the height of the wave from crest to crest, 
or from trough to trough, ^ the length. The wave is assumed 
to be of the form indicated by the " trochoidal theory," but no 
account is taken of the internal structure of the wave. 1 

Construction of a Trochoidal Wave Profile. The wave on 
which a ship is supposed to be momentarily poised has for its 
profile a curve called a " trochoid." This is a curve traced out 
by a point inside a circle when the circle is rolled along a 
straight line. The curve can be traced by its co-ordinates 
referred to axes through the crest (Fig. QQA), viz. 

T $ h a 
x = L . . sin 9 

27T 2 
hi /l\ 

y = -(i cos 0) 

6 being given the values of 30, 60, etc., in circular measure. 
L is the length of wave from crest to crest, and h is the height 
from crest to trough. The curve may also be drawn by the 




FIG. 



construction indicated in Fig. gg\. It is noticed that the 
curve is sharper at the crest than in the trough, which is a 
characteristic feature of sea-waves. 

In order to get the displacement of the ship when on the 

1 See the "Manual of Naval Architecture," by Sir W. H. White, 
chaps, v. and viii. ; and a paper at the Institution of Naval Architects, by 
Mr. (now Sir) W. E. Smith, M.I.N.A., in 1883; see also later. 



Strains experienced by Ships, etc. 



271 



wave, it is convenient at each square station to run in a curve 
of sectional areas. Then, if the profile of the wave is traced 
and put on the profile of the ship, the area of each section up to 
the surface of the wave is at once measured off, and these areas 
integrated throughout the length give the displacement and 
centre of buoyancy. If these are not correct, a further trial must 
be made ; the ship may 



have to be trimmed to 
get the C.B. right, it 
being essential that this 
is in the same section 
as the C.G. 

With these assump- 
tions we can proceed 
to construct the " curve 
of buoyancy" for both 
cases (i) and (2), and 
from it and the u curve 
of weights " we obtain 
the "curve of loads." 
Then, by the principles 
explained above, we can 
determine the " curve of 

shearing force " and the " curve of bending moment." In Fig. 
100 is given a set of curves for a ship on a wave-crest, and in 
Fig. 1 01 a set of curves for the same ship astride a wave-trough. 

In any case the maximum bending moment may be expressed 

in the form wei g ht X len g th an( j j t j s foun d t h a t the value of 
coefficient 

this coefficient will not usually fall below 20 for either of the 
extreme cases taken above. The maximum bending moment 
in foot-tons for ordinary ships may be generally assumed at 
from ^ to the product of the length in feet and the dis- 
placement in tons. The latter is frequently taken as a 
standard value. In regard to this matter Mr. Foster King 
made the following remarks at the I.N.A., 1915: "For 
all ordinary vessels such as those with which we have to 
deal, and many extraordinary ones, it has been found that 
a close approximation to the greatest bending moment 




FIG. 100. 



272 



Theoretical Naval Architecture. 



obtained from direct calculations is obtained from the formula 
-, where L is the length, B the breadth, and 



35 X 35 
D is winter draught in feet. 




FIG. 101. 



Twenty years' experience of the 
application of this for- 
mula has shown an 
agreement of the order 
of about 5 per cent, with 
the calculated figures 
furnished by builders 
for vessels of the most 
widely dissimilar cha- 
racter and dimensions 
and give evidence of its 
utility as a guide in 
strength investigations." 
This maximum bending 
moment will usually 
occur somewhere in the 
vicinity of the midship 
section. 

It is usual to draw the distribution of weight as a series of 
steps. For a large ship the length can be divided into sections 
bounded by the main bulkheads, and the weights in each section 
grouped together and plotted as uniform over each section. 
Where extreme accuracy is desired, the work can be done in 
greater detail. Fig. IOIA gives the curves for a torpedo-boat 
destroyer lying in the trough of a wave of height ~ her length, 
all the bunkers being full. The use of a machine called the 
" integraph " is of great value in getting quickly the curves of 
shearing force and bending moment from the curves of loads. 
If the pointer of this machine is run round the boundary of 
a curvilinear area, the pen traces out a curve the ordinates of 
which give the area of the original up to corresponding points. 
This is just what is required in the above calculations. A 
paper on the integraph was given by Mr. Johnson at the 
Scottish Shipbuilders in 1904, to which the student is referred 
for further information. See also a paper by Professor Biles 
before the I.N.A. in 1905, for an exhaustive discussion on the 



Strains experienced by Ships^ etc. 



273 



strength of ships, with special reference to calculations and 
experiments on H.M.S. Wolf. 

Stress on the Material composing the Section. 
Considering now the ship's structure as a girder, a hogging 




FIG. IOIA. 



moment produces tension in the upper portion of the girder and 
compression in the lower portion of the girder, the reverse being 
true for a sagging moment. 

We now have to consider in some detail how a given beam 

T 



2/4 



Theoretical Naval Architecture. 



is able to withstand the stresses on its material when subjected 
to a given bending moment. Take a beam bent as in Fig. 92 
and Fig. 102. AB is a longitudinal section and LL is a 
transverse section of the beam. The upper layers are shortened 
and the lower layers are lengthened. There must be one inter- 
mediate layer which is unaltered in length. This layer is 




FIG. 



called the " neutral surface," and the transverse section SS is 
called the " neutral axis." This neutral axis can be shown to 
pass through the centre of gravity of the section. 1 The bending 
moment at the section LL is resisted by the compressive 
stresses in the upper layers and the tensile stresses in the 
lower layers. 

It can be shown x that the following relation holds : 

M 



where / is the stress in tons per square inch at distance y 

inches from the neutral axis. 

M is the bending moment at this section in inch-tons. 
I is the moment of inertia of the section about the 

neutral axis in inch-units. 

It is by this formula that the stress on a particular portion 
of the section of a beam can be determined, when we know the 

1 See any standard work on " Applied Mechanics," as that by Professor 
Cotterill, F.R.S. 



Strains experienced by Ships, etc. . 275 

bending moment at that section, the position of the neutral 
axis, and the moment of inertia of the section about the neutral 
axis. 

Take, for example, the various sections of beams in Fig. 93. 
Length of beam 12 feet, beam loaded in the middle with i ton 
(neglecting the weight of the beam). 

For (a) y = i" 

1 = ^x16x4 = $ in inch-units 
M = 36 inch-tons 
/. p = the stress at the top or bottom 

36 X 12 
= I X ^? - 
64 

= 6' 75 tons per square inch 

For (*) y = 2" 

I = ^ in inch-units 
M = 36 inch-tons 

the stress at the top or bottom 




3*375 tons per square inch 



For (<:) y = 4" 

I 2jjfl 



in inch-units 
M = 36 inch-tons 
/ = the stress at the top or bottom 



1*6875 tons P er sc l uare 



For (rf) y = 4 

I = * 



$* in inch- units 
M = 36 inch-tons 
/ = the stress at the top or bottom 



= I '02 tons per square inch 

Or looking at the question from another point of view, if 
we say that the stress on the material is not to exceed 10 tons 
per square inch, then we can determine for each of the sections 
in Fig. 93 the greatest bending moment to which the beam 
can be subjected. 

For (a) M = - xl = ^xf$ = 53g inch-tons 
jr or () M=-Xl = Tj Q X^j <5 = lo6l inch-tons 



276 Theoretical Naval Architecture. 

For (c) M = ^ x I y X 2s = 2I 3 J inch-tons 
For (</) M=^xl = x*i* = 353$ inch-tons 

It is thus seen that the ratio of the bending moments that 
these beams can stand is' 

53 i :io6| : 2i 3 i :.353| 

or 
1:2:4:6!. 

The area of each section is the same, the only difference being 
in the different distribution of the material of the section with 
reference to the neutral axis. 

We come now to the case of a ship subjected at a particular 
section to either a " hogging " moment or a " sagging " moment. 
To determine the stress on any portion of the section, we con- 
sider the vessel to be a large beam subjected to a given 
bending moment, and we apply the formula 



There are two things to be found before we can apply this 
formula to a given section, viz. : 

(i.) The position of the neutral axis, which passes through 
the centre of gravity of the section. 

(ii.) The moment of inertia of the section about the neutral 
axis. 

In considering the strength longitudinally of a section, 
account must be taken only of such material as actually con- 
tributes to the strength through an appreciable length in the 
vicinity of the section, such as plating of the inner and outer 
bottom, keel, continuous longitudinals or keelsons, stringers, 
deck-plating, planking, etc. 

A distinction must be made between material in tension and 
material in compression. In tension, allowance must be made 
for the material taken away from the plating, etc., for the rivet- 
holes, but in compression this deduction is unnecessary. It is 
also usual to consider that wood is equivalent to y^ ^ ts area 
in steel for both tension and compression. For an armoured 



Strains experienced by Ships, etc. 



277 



vessel the armour is not assumed to take any tension, but is 
assumed to be effective against compression. 

We must accordingly have two separate calculations one 
for the section under a hogging moment, and one for the 
section under a sagging moment. The position of the neutral 
axis and the moment of inertia of the section about the neutral 
axis will be different for each case. 

With reference to the above assumption, the following 
remarks of Dr. Bruhn (I.N.A., 1899), wno ^ as given great 
attention to this subject, may be noted : 

Dr. Bruhn thinks that the correction for the rivet-holes in the calcula- 
tion for I is more an act of error than of correction. This correction 
assumes the structure highly discontinuous, the I and the position of the 
neutral axis varying at- the frame and between the frames. But the whole 
theory of bending is based on the assumption that the structure is con- 
tinuous, and the bending certainly must be continuous. The I should 
therefore be taken for the solid section^ and if it is desired to find the stress 
between the rivets, we may increase the stress in the ratio in which the 
sectional area is reduced (say g). This method for other than armoured 
ships only requires one calculation for the moment of inertia, and when 
dealing with shearing stresses, there is no more reason for deducting rivet- 
holes on one side than on the other. 

The following form may conveniently be used for calculating 
the position of the neutral axis and the moment of inertia of the 
section about the neutral axis, areas being in square inches and 
levers in feet : 



I 
Items. 


2 

Effective 
area in 
square 
inches. 


3 

Lever 
in feet. 


4 

Moment. 


5 

Lever 
in feet. 


6 

Moment 
of 
Inertia. 


7 

A X a A x 

h> 
















A 




M 




I 


9 



A section of the ship should be drawn out to scale with all 
the scantlings shown on. An axis is assumed at about ctae-half 
the depth of the section. The several items are entered in 



278 - Theoretical Naval Architecture. 

column i, the effective areas in column 2, and the distances of 
the centres of gravity from the assumed axis in feet are entered 
in column 3. For the items below the axis these levers are 
negative. We thus obtain column 4, which gives the moment 
of each item about the axis, and the algebraic sum of this 
column, M, divided by the addition of column 2, viz. A, gives 
the distance of the neutral axis from the assumed axis, in feet, 
say d feet. 

We now place in column 5 the same levers as in column 3, 
and multiplying the moments in column 4 by the levers in 
column 5, we obtain the areas of the several items multiplied 
respectively by the square of their distances from the neutral 
axis. Each of these products is, of course, positive. All these 
added give a total I, say. For the portions of the section which 
are vertical, an addition is needed for the moment of inertia of 
the items about axes through their own centres of gravity, viz. 
Y5 . A . fi 2 (see p. 104). For portions of the section which are 
horizontal, h is small, and this addition may be neglected. We, 
therefore, arrive at the moment of inertia of the section about 
the assumed axis, viz. I + i" = I A , say. We now have to transfer 
this moment of inertia about the assumed axis from that axis to 
the neutral axis, or I = I A - A X d* y as explained on p. 104. 

We can now determine the stress on the point of the section 
farthest from the neutral axis, as this will be the point at which 
the stress is greatest, by using the formula 

M 



The stresses thus found are only comparative, and must 
be compared with those for a ship on service found to show 
no signs of longitudinal weakness. Large ships can bear a 
stress of 10 tons per square inch, because the standard wave is 
exceptional. Ships 300 to 400 feet long have stresses 6 to 7 
tons per square inch. In the special case of the Lusitania^ 
a stress of 10 tons per square inch was allowed when on a 
wave of her own length of height one-twentieth the length. 
It is to be observed that such a wave is of quite phenomena) 
size, and unlikely to be encountered. 



Strains experienced by Ships, etc. 



279 



Specimen Calculation for the Moment of Inertia, 
etc., of Section. The following specimen calculation for a 
torpedo-boat destroyer is given as a guide to similar cal- 
culations. The depth of the girder was 17*7 feet and the 
assumed neutral axis was taken as 9 feet above underside of 
flat keel. 

The column for areas was filled in with weight per foot run, 
the total being turned into areas at the end (/ (area) means, 
proportional to area, i.e. a function of area). 

Above assumed neutral axis. 



One side only. 


W 

/(Area) 
Ibs. per ft. 


(2) 

Lever. 


(3) 
/(Moment). 


(4) 
/(I), 

i.e. ( 2 )X( 3 ) 


(5) 
d 


(6) 


Deck stringer . 


67*5 


8-28 


559 


4630 






Remainder deck . 


74*38 


8-61 


640 


55 10 






Girder coaming . 


I4-0 


8'95 


123-6 


II2I 


I'7S 


43 


Girder upper 














angles . . . 


8-0 


9*24 


74-0 


68 3 






Girder deck angles 


3-0 


8-84 


26-5 


234 






Girder lower 














angles . . . 


7' 


8-06 


56-4 


455 






2nd girder plate . 
2nd girder angles 


4*2 

8-0 


8-17 
8-17 


} 99-6 


813 






3rd girder plate . 
3rd girder angles 


4-2 
8-0 


8-07 
8-07 


} 98*4 


793 






Bunker bulkhead 














top plate . . 


24*0 


6-7 


160-7 


1079 


3*o 


216 


Bunker bulkhead 














2nd plate . . 


i6'8 


3*92 


65*9 


258-5 


2-8 


132 


Bunker bulkhead 














3rd plate . . 


17-1 


1-22 


20-8 


25*5 


2-85 


139 


Gunwale angle . 


7*o 


8-00 


56-0 


448 






Sheer strake . . 


63*0 


5'8o 


365*3 


2120 


4*5 


1275 


Strake below . . 


40-95 


1-58 


64*7 


102 


4*55 


847 


Side stringer . . 


8-97 


3*03 


27-2 


82 






376-1 2438-1 18,354 12)2652 

221 - 


iS.ot; 221 



fl for tension 307-9 



1995 15,205 



280 Theoretical Naval Architecture. 

Below assumed neutral axis. 



One side only. 


AArea) 
Ibs. per ft. 


Lever. 


/(Moment). 


/(I) 


d 


y(A.</). 


D strake . 


3672 


2-39 


877 


210 


3-8 


53 J 


C strake . 


46*0 


5-55 


255 


I42O 


2-9 


387 


B strake . 


44'55 


7-62 


340 


2590 


i"4 


87 


A strake . 


55-62 


8-58 


477 


4095 


0-8 


35 


i Flat keel 


30-0 


9*0 


270 


2430 






} Vertical keel 


9-0 


8-2 


7T8 


60 5 


i '5 


20 


Lower angles 


7-0 


8-87 


62-1 


550 






\ Upper angles 
I Rider plate 


6-0 

8-33 


7-52 
7-42 


6r8 


339 

457 






ist long, angles 


6-0 


7 '4 


44'4 


328 






2nd long, plates 


5-36 


6-9 


37 'o 


255 


0-67 


2 


2nd long, angle 


3-0 


7-2 


21-6 


155 






2nd long, angle 


5-0 


6.6 


33"0 


218 






Lower side 














stringer . . 


8-97 


i '37 


12-3 


17 






Bilge keel plates 


I0'0 


6-6 


66-0 


435 


0-8 


6 


Bilge keel angles 


II'O 


6-25 


687 


430 






Bunker bulkhead 


16-8 


i '47 


247 


36 


2-8 


*32 


Bunker bulkhead 


32*0 


473 


I5i'5 


717 


4-0 


5 12 




34I-35 




2131-7 


15,288 




12)1711 









- 


143 















I5.43I 




'43 



for tension 279 1743 12,630 

Hogging. For hogging the full area below axis is taken, 
and T 9 T the full area above axis to allow for rivet holes 
/(Area) = 307-9 + 34i'3 = 649*2 

area = X 649-2 = 381-9 sq. in. 

N.A. below assumed! _ 2131-7 1995 

axis J 649-2 

/(I)aboutassumedaxis= 15,205 4- 15,431 = 30,636 



0*2 1 ft. 



I about N.A. = - - [30,636 - (649-2 x 0-2 1 2 )] = 18,000 

3'4 

N.A. above keel = 879 ft. N.A. below deck = 8-91 ft. 
Bending moment = 10,275 ft. tons 

Tensile stress in deck = -~ - x 8*91 = 5-1 tons sq. in. 
10,000 

Compressive stressing 10,275 

} - - x 879 = 5-02 tons sq. in. 
keel J 18,000 

Sagging. For sagging the full area above axis is taken, and 
the full area below axis to allow for rivet holes. 



Strains experienced by Ships, etc. 281 

/(Area) = 376-1 + 279 = 655-1 
area = -^X 655-1 = 385*4 

J T" 

N.A. above assumed! __ 2438 - 1743 __ ro6 ft 

axis J 655-1 

/(I) about assumed axis = 18,575 + 12,630 = 31,205 

I about N.A. = -4- [31,205 - 6 55 >]C X i'o6 2 ] = i7,9 20 

o 4 

N.A. above keel = 10*06 ft. N.A. below deck = 7-64 ft. 
Bending moment = 13,000 ft. -tons 

Tensile stress in keel = ~ X 10-06 = 7-3 tons sq. in. 



Oppressive stress in) 13,000 fi = tons ^ 
deck J 17,920 

Equivalent Girder. Although not necessary for calcu- 
lation purposes, it is frequently the practice to draw out for 



HOGING. -- 




FIG. 103. FIG. 104. 

the case under consideration a diagrammatic representation of 
the disposition of the material forming the section. Such a 
diagram will show at once how the material is disposed relative 
to the neutral axis, and gives the section of the girder that 
the ship is supposed to be. Such a diagram is termed the 
" equivalent girder," and there must be one for hogging and 



282 



Theoretical Naval Architecture. 



one for sagging, as shown in Fig. 103 and Fig. 104 respectively, 
which are the equivalent girders for an armoured battleship. 

A number of examples of equivalent girders for merchant 
ships are given in Mr. Foster King's paper, I.N.A., 1913. 

Shearing Stresses. We have seen above that to dispose 
the material of a ship so as to resist most effectively the stresses 
due to a bending moment, we must pay special attention to 
the upper and lower portions of the girder. There are, how- 
ever, also shearing stresses in a loaded structure which, under 
certain circumstances, may cause straining action to take place. 
Professor Jenkins called attention to these stresses in a paper 
before the I.N.A. in 1890, and their importance has been 
increased in recent years owing to the great increase in the 
size of the vessels built. 

It is necessary first to deal with the shearing stresses which 
occur in an ordinary loaded beam. In a beam, besides the 
bending moment at each section, there is a tendency for each 
section to slide over the adjacent one. This is measured by 
the " shearing force." At each point of the section there is a 
shearing stress set up to resist this sliding tendency. It can be 
shown that such shearing stress is always accompanied by a 
shearing stress of equal intensity on a plane at right angles. 

Consider two consecutive sections of a beam K' and K" Sx apart 
(Fig. IO4A), at which the^bending moments are M and M + 5M respec- 

K' K" 





FIG. 1 04 A. 



tively. Then, if/ is the normal stress at section K' at a distance y from 
the neutral axis, and 5A is the element of area on which this stress acts, 
the total normal force on the section above AB is 



VI I 

on proceeding to the limit. If now m is the moment of the area above AB 



Strains experienced by Ships, etc. 283 

about the neutral axis, the normal force is y m. At the consecutive 

section this will be . m. The difference between these is the 
resultant horizontal force on the portion of the beam 5* long above AB, 
viz. . 5M. This must be the shearing force causing the section above 
CD to slide along. If, now, q is the intensity of the shearing stress 
along CD, and b the breadth of AB, we have q . b . 5x .- . SM, or 



in th j ^ 

ax L . b dx 

the shearing force on whole section, so that we have q ' . This 

shearing stress is therefore zero at the top and bottom of the section, and 
it will vary at other points of the section. 

To illustrate the variation of this shearing stress, take a beam of I 
section or hollow section, as Fig. IO4A, subjected to a total shearing force 
of 10 tons. It will be found that the variation of the shearing stress is 
represented by the right-hand portion of figure. It takes a sudden jump 
at bottom of flange from 0*25 to 1*24 tons per square inch, because the 
breadth is suddenly diminished from 5 inches to I inch. The maximum 
value is 1*56 tons at the neutral axis. It is thus possible, in a beam with 
a thin web, that an excessive shearing stress may be set up, and just at 
that part of the section where there is no stress due to the bending moment. 
It is also to be noticed that the shearing strength of steel is about four- fifths 
the tensile strength. 

In a ship the shearing force amidships is usually zero (see 
Figs. 100, 101), and the shearing force reaches a maximum at 
about a quarter the length from each end. This, therefore, 
will be the portion at which shearing strains are likely to be 
most severe, and the maximum strains will occur in the neigh- 
bourhood of the neutral axis, because here the breadth of the 
section is usually only twice the thickness of the bottom plating. 
This stress has shown itself by the working of the rivets in 
these portions of large ships, and the stresses vary also in 
opposite directions according as the ship is in the trough or 
on the crest of a wave. It is, therefore, becoming the practice 
to work treble-riveted fore-and-aft laps at about mid-depth in 
the fore and after bodies of large ships. 1 

There are also set-up stresses due to bending of the plating 
owing to the varying pressure of water, and these, together with 

1 Lloyd's Rules say: "In vessels of 480 feet and upwards, with side 
plating less than 0*84 inch in thickness, the landing edges are to be treble- 
riveted for one-fourth the vessel's length in the fore and after bodies for a 
depth of one-third the depth." 



284 Theoretical Naval Architecture. 

the above stresses, have shown themselves by working in the 
parts above mentioned, which has required the special strength- 
ening referred to. (The subject has been exhaustively discussed 
by Dr. Bruhn in a paper before the Scottish Institution of Ship- 
builders, 1902.) 

Principal Stress. When the material of a beam is 
subjected to a tensile or compressive stress, together with a 
shearing stress, these combine together to produce what is 
termed the " principal stress " at any particular place. It can 
be shown that if / is the ordinary tensile or compressive stress, 
and q is the shearing stress, then the principal stress/ is given 
by the equation 



Thus, take a place immediately beneath the deck of a ship 
with the stern overhanging in dry dock; A = 3' T > ^ = x '4 2 - 
Then/ = 3*65 tons, which is seen to be greater than the simple 
tensile stress. 

Unsymmetrical Bending. In the ordinary investiga- 
tion we assume that the ship is upright. If a ship is inclined 
the depth of section is increased, and it may possibly happen 
that an increased stress would be experienced at the corner of 
the section. 

Let MM (Fig. 1043) be the axis of the bending moment, 
the ship being heeled to an angle 0. Then M, the bending 





FIG. 1048. 

moment, may be resolved into M . cos 6 with the axis OX, and 
M . sin 6 with the axis OY, O being the C.G. of section. Each 



Strains experienced by Skips, etc. 285 

of these will produce stress at P as if it acted alone, and 
the total stress at P will be 



I 1} I 2 being the moments of inertia about axes OX, OY. The 

position of the neutral axis is where / = o, or where - = - 1 

x 1 3 

X tan 6 = tan <f>. If the point farthest from NN has co-ordinates 
y and x' referred to OX, OY, then the maximum stress is 

x' \ 

^ . cos 6 + =- . sin 6 } 

AI 2 

Professor Biles (Scottish Shipbuilders, 1893-4) gives the 
results for a ship for all angles from o to 90. He found, for 
the ship he took, that the maximum stress was reached at 30, 
and was there 20 per cent, greater than when the ship was 
upright. 

The " Smith " correction due to taking account 
of the Internal Structure of a Wave. 1 In order to 
understand this it will be necessary to deal with some features 
of the trochoidal wave theory. 

A trochoid is a curve traced out by a point inside a circle 
when the circle is rolled along a straight line. Its co-ordinates 
relative to axes through the crest are (Fig. 99A) 

_ k u H /}! 

x ~ 27r ~ 7 S1 1 I where L and H are the length 

H I and height of the trochoid 

y= -(i -cos0) } 

If we imagine the circle rolled along with a velocity v, and 
then a backward velocity v impressed on the system, we have 
the points at extremities of tracing arms moving with uniform 
angular velocity, and the wave formation will have a velocity 
v. Fig. 1040 shows how the wave travels due to the revolu- 
tion anti-clockwise of the points P. The particles in the crest 
move in the same direction as the wave advance, and in the 
trough in the opposite direction. A wave is the passage of 
motion, and there is only a relatively small actual motion in 
circular orbits of the particles of water composing the wave. 
1 See a paper by Mr. (now Sir) W. E. Smith, I.N.A. 1883. 



286 



Theoretical Naval Architecture. 



If R be the radius of the rolling circle then L = 27rR, and 
= g . (dimensions in feet and seconds). The line of 




FIG. 1040. 

orbit centres is above the level of still water an amount J . 

' R 

where r is the radius OP for the surface trochoid or J . H (see 
Fig. 1040). 

For sub-surfaces (see Fig. io4E) the rolling circle is the 




FIG. 1040. 

same. The axis is above the level of the same particles in 

still water an amount i . - where r is the radius for the sub- 
j\. 

surface in question. 

For a sub-surface the centre of whose rolling circle is at a 
distance y below that of the surface trochoid, we have 

r = r . e~&, where e is the base of Napierian logarithms =2718. 
The values of r for values of y = i, 2, 3, etc. are therefore 



_ 

(y = 2), r 2 = r . e R 
(y = 3), r, = r Q .<rl' etc ' 
To evaluate these, we take logarithms first to the base 

log. n = log. r - R , since log, e = i 
log. r a = l og. r ^, and so on. 



Strains experienced by Ships, etc. 



28; 




288 Theoretical Naval Architecture. 

The pressure at any point in a trochoidal wave is the same 
as at the point it occupies when in still water. Thus in Fig. io4E, 
along the sub-surface BB, the pressure is the same as at its still- 
water level b, and not due to its distance below the surface 
trochoid (as is assumed in the standard method of calculation). 

We therefore draw the trochoidal profile of the wave we 
have to deal with and also the sub-surfaces corresponding to 
lines of orbit centres at distances say 2 feet apart. We then 
calculate the positions of the corresponding still-water levels, 
a, b, c, d, <?,/. 

Take as an illustrative example the wave drawn in Fig. io4E, 
60 feet long, and 1 2 feet high. (This is, of course, an exagge- 
rated ratio of H 4- L, but has been selected for the sake of 
clearness). The lines of orbit centres have been taken at 
2 feet intervals, and the profile of the surface trochoid can be 

drawn as already described In this case R = = 9-55, 

and r = 6, being one-half of the height of the surface trochoid. 
In order to draw the sub-surface trochoids we have to find the 
radii for values of y = 2, 4, 6, etc. 

We have log, r = log, r Q ^ 

Turning into ordinary logarithms to the base 10 (See Appendix 
B.) we have 

2*3 R 

Putting in values of r and R, and successive values of y = 2, 
4, 6, 8, 10, we have for values of r\ 4-86, 3-94, 3*19, 2'6 and 
2' i. These will be the half heights of the sub-surface trochoids 
which can then be drawn in as indicated in the figure. The 
level of still water below the lines of orbit centres is then 

obtained by putting in the values of r in the expression , 

2K 

or r88, 1-24, o'8i, 0-53, 0-35, 0*23, which are set down at 
the side of the figure giving the levels a, b, c, d, <?, /. Thus 
the pressure at any point on the sub-surface DD is that due 
to the distance of d below a. At the crest, if the wave pres- 
sures were neglected, the pressure at D would be that due to 



Strains experienced by Ships, etc. 289 

a head of 8*8 feet, whereas really it is only a pressure due to 
a head of 47 feet. 

Take now a ship on the wave and consider the section 
that comes at HH. The levels of the surface A, and the sub- 
surfaces B, C, D, etc., are placed on the section (Fig. io4F), 
A, B, C, etc. At the level of B we set up bb' = ab, at fc he 
level of C, cc' = ac> etc., and a curve through ab'c', etc., will 
give the line to work to to obtain the true value of the 
buoyancy due to the section. In the case of sections in the 
crest portion this is less than that up to the level of the wave. 
Similarly for a section as at EE (Fig. 1040), where the value 
of the effective buoyancy is greater. This done for sections 
all along the length will result in a curve of effective areas. 
When the ship is on the crest this curve is as dotted in 
(Fig. io4H), and in the trough as dotted in (Fig. 104;), as 
compared with the full curves obtained in the standard method. 
These new curves of buoyancy must, of course, satisfy the 
ordinary conditions, viz. that the displacement and position 
of the centre of buoyancy are correct. 

Trochoidal Wave Theory. The following are the 
principal formulae arising out of the trochoidal wave theory, 
where 

L is length of wave in feet (crest to crest). 

H is height of wave in feet. 

T is periodic time in seconds, i.e. time of traversing the 
length. 

V is velocity of wave in knots (i.e. 6080 feet per hour). 

v is velocity of wave in feet per second. 

R is radius of rolling circle in feet. 

;- is radius of tracing arm in feet, i.e. one-half the height 
of wave = H. 

r is radius of tracing arm in feet at depth of y where 

y is depth of line of orbit centres from that line for the 
surface trochoid. 

g is acceleration due to gravity, viz. 32-2 in foot second 
units. 

V Q is a velocity of a particle in its orbit in feet per 
second. 



Theoretical Naval Architecture. 




Section, E-E 

FIG. 1040. 




FIG. io4H. 




FIG. T<>4j. 



Strains experienced by Ships^ etc. 291 

r = ^- *=*-. R = ^.L = 5-1 2L 

V 2 =rSL, since V = z,x 



r r d e~K, where e = 2718 log e e = i 

r = Iog 10 r Q 'jp7 

(since log, A = 2-3 Iog 10 . A). 



log, r = log, r - j|, and log, r = Iog 10 r Q 'jp7' g 



2vrr TT.H H 

v Q = -y- . v # for surface particle = -y X # = 7*1 -7^ 



Height of centre of orbits of a given particle above the level 
of that particle in still water = ^ . For the wave surface 

7T.H 2 

this distance = ^- . 

R-r 

Virtual gravity at crest = 5- . . 

R + r 
at trough = . g. 



EXAMPLES TO CHAPTER VII. 

1. Determine the maximum stress on the section of an iron bar, 2 
inches square and 20 feet long, when supported at the ends and unloaded 
with one side horizontal. Ans. 6000 Ibs. per square inch. 

2. An iron bar of the same length, and supported as in the previous 
question, is of circular section, 2 inches diameter. Determine the maximum 
stress. Ans. 8000 Ibs. per square inch. 

3. A vessel floating in still water is subjected at a certain section to a 
bending moment of 144 foot-tons. Determine the longitudinal stresses 
(in pounds per square inch) in the material at top and bottom of this 
section, assuming the section to be rectangular, 21 feet wide, 10 feet deep, 
J" thick, and that the whole of it is effective in resisting stresses. 

Ans. 223 Ibs. 

4. The buoyancy of a vessel is o at the ends and increases uniformly to 
the centre, while the weight is o at the centre and increases uniformly to 
the ends. Draw the curves of shearing force and bending moment, and 
find the maximum values of these quantities in terms of the displacement 
and length of the vessel. 

Ans. \ W, & W.L. 

5. Suppose the skin and plate deck of an iron vessel to have the 
following dimensions at the midship section, measured at the middle of 
the thickness of the plates. Find the position of the neutral axis and 
moment of resistance to bending. Breadth 48', and total depth 24', the 



292 Theoretical Naval Architecture. 

bilges being quadrants of 12' radius. Thickness of plate g" all round, and 
coefficient of strength / = 4 tons. 

Ans. Neutral axis 13" above centre of depth. 
Moment of resistance to hogging, 32,500 foot-tons. 

i> sagging, 39,000 

(Examples 4 and 5 are from "Applied Mechanics," by Professor 
Cotterill, F.R.S.) 

6. A ship on a wave-crest is subjected at the midship section to a 
hogging moment of 28,000 foot-tons. The depth of the section is 37-5 
feet, and the neutral axis is 18*2 feet from the bottom, the moment of 
inertia of the section about the neutral axis is 477,778 (square inches 
X feet 2 ). Determine the maximum compressive and tensile stresses. 

Ans. 1*07 tons per square inch compressive at bottom of section. 
1*13 tons per square inch tensile at upper part of section. 

7. State the maximum bending moment (in terms of weight and length 
of vessel) in the case of a vessel having the weight uniformly distributed 
and the curve of buoyancy a parabola. State also the position where 
these maxima occur. 

W 

Ans. S.F. at o'2iL from end = . 

10-5 

B.M. amidships = 

8. In question 3, if the ship has a bulwark each side 2 feet high, J inch 
thick, what will then be the maximum stress ? Explain the significance 
of your result as applying to actual ships. Ans. 263 Ibs. 



Increase of depth of section will not necessarily diminish the maximum 
stress. 



. 

y I p y I 

and 8/ will be negative, i.e. stress diminishes only if > ^ . 

This point acquires special importance 'in vessels with a light con- 
tinuous superstructure, as 

(1) Boat deck in large cruisers. 

(2) Superstructure in merchant vessels. 

If the structure is made continuous, it is found that the influence of the 
increased depth is greater than the increased I, and thus greater stresses 
are likely to be experienced by the superstructure than it can bear. For 
this reason, either 

(1) A sliding joint is made, so that the superstructure contributes 

nothing to the structural strength ; or, preferably 

(2) The superstructure is made an integral part of the ship's structure. 

See a paper by Mr. Montgomerie, I.N.A., 1915. 
It does not follow that material added to a section will diminish the 
maximum stress. We have 

M 9y 51 



Suppose a small area a is added at a distance /from neutral axis, then 
this axis will shift : 

JLiL 

A+a 



Strains experienced by Ships, etc. 293 

The new I about old axis = 1 + a ./- 

The new I about new axis = I + a ./ 2 - (A + a}( j^-^ J 

... increment of I or 51 = a ./> - 



_ af 



and I = A . # 
S/ is positive if l - > - or - 

/.*. the stress increases at distance y if the added material is placed less 

& 
than from the neutral axis. 

/ 

In a rectangular beam, if material be added less than \ the depth 
from mid-depth the stress is increased. 

If a square inches of plating, placed at a distance of h feet above the 

top of the girder, is such as to give the san,e stress as before (i.e. -is the 
same), show that 

S.-A.J 



A(y + h)* + I 

This formula was used by Mr. Montgomerie in his paper before 
I.N.A., 1915, on "The Scantlings of Light Superstructures." 

New I _ I 

New_y y 

9. A rectangular vessel is 30 feet broad and 20 feet deep, and has deck 
plating \ inch thick, sides and bottom inch. At a certain section it is 
subjected to a hogging moment of 20,000 tons-feet, and a shearing force 
of 300 tons. Calculate in tons per square inch 

(1) Maximum tensile and compressive stresses. 

(2) Maximum shearing stress. 

(3) Principal stress immediately under deck. 

Ans. (i) 7-1, 4-9 ; (2) 1-42 ; (3) 7-2. 

10. In the previous example, what would be the stresses if the deck 
and bottom were J inch thick, and the sides \ inch ? 

Ans. (i) 5, 5; (2) 2-62; (3) 5-86. 

11. An I beam 8 inches deep and I inch thick, with flanges 5 inches 
wide, overhangs a distance of 5 feet, and a weight of 5 tons is placed at 
the end. Determine at the point of leaving support (tons per square inch) 

(1) Maximum tensile and compressive stresses. 

(2) Maximum shearing stress. 

(3) Principal stress immediately below the upper flange. 

Ans. (i) 8-5, 8-5; (2) 078; (3) 6-46. 

12. In a vessel of 10,000 tons displacement and 450 feet long, the 
maximum bending moment is 3 \jW.L. The depth of midship section is 
39 feet, and the neutral axis for hogging is at 0*49 the depth from keel. 



294 



Theoretical Naval Architecture. 



The I for hogging about the neutral axis is 4000 in foot units. Calculate 
the maximum tensile and compressive stresses. 

Ans. 5'i8, 4*97 tons per square inch. 

13. A vessel of 3000 tons displacement and 360 feet long has a maximum 
hogging moment of ^ W.L. The draught is 14^ feet, and the freeboard 
to stringer is 7$ feet. The neutral axis for hogging is 3*15 feet below the 
water-line, and the moment of inertia about neutral axis is 73,000 (square 
inches X feet 2 ). What are the maximum tensile and compressive stresses ? 

Ans. 5'i8, 5*41 tons per square inch. 

14. A vessel is designed with a very large overhang at the stern from 
the cut-up of the keel. Indicate what calculations you would make to see 
if the ship could safely be dry docked with the stern unsupported. Indi- 
cate how you would strengthen such a ship to withstand the strains set up 
in dry dock. 

15. Two similar vessels are respectively 300 feet long, 2135 tons dis- 
placement, and 360 feet long, 3000 tons displacement, the depth being 
nearly the same. Indicate the method you would adopt to ensure the 
second ship being strong enough, and estimate the increase of maximum 
bending moment the second ship has to stand as compared with the first. 

Ans. About 68 per cent, greater B.M. 

1 6. The effective part of the transverse section of a vessel amidships is 
represented by the diagram, the vessel being 42 feet broad and 28 feet deep. 

Find the maximum tensile and 

; compressive stresses when the vessel 

is subjected to a sagging moment of 
60,000 foot-tons. The plating is 
\ inch thick and no allowance need 
be made for rivet holes and laps of 
plating. (Honours B. of E., 1908.) 
This example is worked out 
below. The best way to proceed is 
to prepare a table similar to that in 
this chapter ; attention is necessary to 
the units, the areas being in square 
inches and the lengths in feet. We 
therefore have, taking in the first 
place all distances from the keel 



t 



Items. 


Area in 
sq. in. 


Levers 
in ft. 


Moment. 


Levers 
in ft. 


Moment of 
inertia. 


A.A.* 


Upper deck 
Main deck . 


252 
252 


28 
2O 


7,056 
5>40 


28 
20 


197,578 
100,800 





Tank top . 


252 


4 


1, 008 


4 


4,032 





Bottom 


252 

















Sides . . . 
Girders . . 


336 
48 


M 

2 


4,704 
9 6 


14 

2 


65,856 

I 9 2 


21,952 
6 4 




1392 




17,904 




368,458 


22,016 



Strains experienced by Ships, etc. 295 

368,458 
17904 _ I2 . 86 ft c G from keel> 22,016 

*39 2 390,474 I about keel in sq. in x ft.* 

I about neutral axis = 390, 474 - 1392 x (i2'86) 2 = 160,224 
stress at keel tensile = X 1 2 '86 4 '8 1 tons sq. in, 

stress at deck compressive = l6o x I S' l 4 = 5' 6 7 ton s sq. in. 

17. An Atlantic ocean wave is 600 feet long and 40 feet high. Cal- 
culate the radii of the orbits at depths of loo, 200, 300, 400, 500 and 
600 feet. 

Ans. 7-03, 2-49, 0*87, 0*31, o'li, 0*04 feet. 

These results show that even in a wave of large dimensions at a depth 
less than the length of the wave the motion of the water is practically nil. 

1 8. The successive crests of the wave profile along a ship's side going 
at speed in still water are observed to be about 300 feet apart. What is 
the speed of the vessel in knots ? 

Ans. 23 knots about. 

19. What is the speed of a 600 feet wave in knots ? 

Ans. 33 knots nearly. 

20. What is the length of a wave successive crests of which are observed 
to pass a stationary observer at intervals of 8 seconds ? 

Ans. 330 feet. 

21. A wave is 600 feet long and 40 feet high. Compare the orbital 
velocity of the particles in the surface with the speed of the wave. 

Ans. 1 1 '6 : 55 '4, or about \. 

22. What is the virtual force of gravity in the crest and trough respec- 
tively of a wave 600 feet long and 40 feet high ? 

Ans. 0.792, I'l ig. 



CHAPTER VIII. 

HORSE-POWER, EFFECTIVE AND INDICATED RESIST- 
ANCE OP SHIPS COEFFICIENTS OF SPEED LA W OF 
COMPARISON PROPULSION. 

Horse-power. We have in Chapter V. defined the " work " 
done by a force as being the product of the force and the 
distance through which the force acts. Into the conception 
of work the question of time does not enter at all, whereas 
" power " involves not only work, but also the time in which 
the work is done. The unit of power is a "horse-power? 
which is taken as " 33,000 foot-lbs. of work performed in i 
minute? or "550 foot-lbs. of work performed in i second? 
Thus, if during i minute a force of i Ib. acts through 33,000 
feet, the same power will be exerted as if a force of 33 Ibs. acts 
through 1000 feet during i minute, or if 50 Ibs. acts through 
ii feet during i second. Each of these will be equivalent to 
i horse-power. The power of a locomotive is a familiar in- 
stance. In this case the work performed by the locomotive 
if the train is moving at a uniform speed is employed in 
overcoming the various resistances, such as the friction of the 
wheels on the track, the resistance of the air, etc. If we 
know the amount of this resistance, and also the speed of the 
train, we can determine the horse-power exerted by the loco- 
motive. The following example will illustrate this point : 

If the mass of a train is 150 tons, and the resistance to its motion 
arising from the air, friction, etc., amount to 16 Ibs. weight per ton when 
the train is going at the rate of 60 miles per hour on a level plain, find the 
horse-power of the engine which can just keep it going at that rate. 
Resistance to onward motion = 150 X 16 
= 2400 Ibs. 

Speed in feet per minute = 5280 
Work done per minute = 2400 X 5280 foot-lbs. 

2400 x 5280 

Horse-power = 

33000 

= 384 



Horse-power, Effective and Indicated, etc. 297 

In any general case, if 

R = resistance to motion in pounds ; 

v = velocity in feet per minute ; 

V = velocity in knots (a velocity of i knot is 6080 feet 

per hour) ; 
then 

1? ^^ <7J 

Horse-power = 

33000 

X V) nearly 



The case of the propulsion of a vessel by her own engines 
is much more complicated than the question considered above 
of a train being drawn along a level plain by a locomotive. 
We must first take the case of a vessel being towed through 
the water by another vessel. Here we have the resistances 
offered by the water to the towed vessel overcome by the strain 
in the tow-rope. In some experiments on H.M.S. Greyhound 
by the late Mr. Froude, which will be described later, the tow- 
rope strain was actually measured, the speed being recorded 
at the same time. Knowing these, the horse-power necessary 
to overcome the resistance can be at once determined. For 
example 

At a speed of 1017 feet per minute, the tow-rope strain was 10,770 Ibg. 
Find the horse-power necessary to overcome the resistance. 

Work done per minute = 10,770 X 1017 foot-lbs. 

TT 10770 X 1017 

Horse-power = 

33000 

= 332 

Effective Horse-power. The effective horse-power of 
a vessel at a given speed is the horse-power required to over- 
come the various resistances to the vessel's progress at that 
speed. It may be described as the horse-power usefully 
employed, and is sometimes termed the " tow-rope " or " tug " 
horse-power, because this is the power that would have to be 
transmitted through the tow-rope if the vessel were towed 
through the water at the given speed. Effective horse-power 
is often written E.H.P. We shall see later that the E.H.P. is 
entirely different to the Indicated Horse-power (written I.H.P.), 



298 Theoretical Naval Architecture. 

which is the horse-power actually measured at the vessel's 
engines. 

Example. Find the horse-power which must be transmitted through 
a tow-rope in order to tow a vessel at the rate of 16 knots, the resistance to 
the ship's motion at that speed being equal to a weight of 50 tons. 

Ans. 5503 H.P. 

Experiments with H.M.S. "Greyhound," by the 
late Mr. William Fronde, P.R.S. These experiments 
took place at Portsmouth as long ago as 1871, and they settled 
a number of points in connection with the resistance and pro- 
pulsion of ships, about which, up to that time, little was known. 
The thoroughness with which the experiments were carried 
out, and the complete analysis of the results that was given, 
make them very valuable ; and students of the subject would 
do well to consult the original paper in the Transactions of the 
Institution of Naval Architects for 1874. A summary of the 
experiments, including a comparison with Rankine's "Aug- 
mented Surface Theory of Resistance," will be found in vol. ill 
of Naval Science. Mr. Froude's report to the Admiralty was 
published in Engineering, May i, 1874. 

The Greyhound was a ship 172' 6" in length between per- 
pendiculars, and 33' 2" extreme breadth, the deepest draught 
during the experiments being 1 3' 9" mean. The displacement 





FIG. 105. 

corresponding to this mean draught being 1161 tons; area of 
midship section, 339 square feet ; area of immersed surface, 
7540 square feet. The Greyhound was towed by H.M.S. 
Active. It was essential to the accuracy of the experiments 
that the Greyhound should proceed through undisturbed water, 
and to avoid using an exceedingly long tow-rope a boom was 
rigged out from the side of the Active to take the tow-rope (see 
Fig. 105). By this means the Greyhound proceeded through 



Horse-power, Effective and Indicated, etc. 299 

water that had not been influenced by the wake of the Active. 
The length of the boom on the Active was 45 feet, and the length 
of the tow-rope was such that the Greyhound's bow was 190 
feet clear of the Aciive's stern. The actual force on the tow- 
rope at its extremity was not required, but the " horizontal 
component." This would be the force that was overcoming 
the resistance, the "vertical component" being due to the 
weight of the tow-rope. The horizontal force on the tow-rope 
and the speed were automatically recorded on a sheet of paper 
carried on a revolving cylinder. For details of the methods 
employed and the apparatus used, the student is referred to 




FIG. 106. 

the sources mentioned above. The horizontal force on the 
tow-rope was equal to the nett resistance of the Greyhound, 
The results can be represented graphically by a curve, abscissae 
representing speed, and ordinates representing the resistance 
in pounds. Such a curve is given by A in Fig. 106. 

It will be seen that the resistance increases much more 
rapidly at the higher than at the lower speeds; thus, on 
increasing the speed from 7 to 8 knots, an extra resistance 
of 1500 Ibs. has to be overcome, while to increase the speed 



3OO Theoretical Naval Architecture. 

from ii to 12 knots, an extra resistance of 6000 Ibs. must 
be overcome. Beyond 12 knots the shape of the curve 
indicates that the resistance increases very rapidly indeed. 
Now, the ratt at which the resistance increases as the speed 
increases is a very important matter. (We are only concerned 
now with the total resistance.) Up to 8 knots it was found 
that the resistance was proportional to the square of the speed ; 
that is to say, if R 1} R, represent the resistances at speeds 
Vj, V 2 respectively, then, if the resistance is proportional to 
the square of the speed 

R, : R 2 : : V, 2 : V a ' 



By measuring ordinates of the curve in Fig. 106, say at 5 and 6 
knots, this will be found to be very nearly the case. As the 
speed increases above 8 knots, the resistance increases much 
more rapidly than would be given by the above ; and between 
1 1 and 1 2 knots, the resistance is very nearly proportional to 
the fourth power of the speed. 

The experiments were also conducted at two displacements 
less than 1161 tons, viz. at 1050 tons and 938 tons. It was 
found that differences in resistance, due to differences of 
immersion, depended, not on changes of area of midship 
section or on changes of displacement, but rather on changes 
in the area of wetted surface. Thus for a reduction of 19^ 
per cent, in the displacement, corresponding to a reduction of 
area of midship section of 16^ per cent., and area of immersed 
surface of 8 per cent., the reduction in resistance was about 
10^ per cent., this being for speeds between 8 and 12 knots. 

Ratio between Effective Horse-power and Indi- 
cated Horse-power. We have already seen that, the 
resistance of the Greyhound at certain speeds being deter- 
mined, it is possible to determine at once the E.H.P. at 
those speeds. Now, the horse-power actually developed by the 
Greyhound's own engines, or the " indicated horse-power " 
(I.H.P.), when proceeding on the measured mile, was observed 
on a separate series of trials, and tabulated. The ratio of the 



Horse-power, Effective and Indicated, etc. 301 



E.H.P. to the I.H.P. was then calculated for different speeds, 
and it was found that E.H.P. 4- I.H.P. in the best case was 
only 0*42 ; that is to say, as much as 58 per cent, of the power 
was employed in doing work other than overcoming the actual 
resistance of the ship. This was a very important result, and 
led Mr. Froude to make further investigations in order to 
determine the cause of this waste of power, and to see whether 
it was possible to lessen it. 

Tf TT T> 

The ratio y Vj ' ' at any given speed is termed the "pro- 
pulsive coefficient" at that speed. As we saw above, in the 
most efficient case, in the trials of the " Greyhound" this co- 
efficient was 42 per cent. For modern vessels with fine lines a 
propulsive coefficient of 50 per cent, may be expected, if the 
engines are working efficiently and the propeller is suitable. 
In special cases, with extremely fine forms and fast-running 
engines, the coefficient rises higher than this. These values only 
hold good for the maximum speed for which the vessel is 
designed ; for lower speeds the coefficient becomes smaller. 
The following table gives some results as given by Mr. Froude. 
The Mutine was a sister-ship to the Greyhound^ and she had 
also been run upon the measured mile at the same draught and 
trim as the Greyhound. 





1 


JH 


g 


i . 


- 




1 


T3 O hn 

8<s s-a 


i 





1 




c 


1'* B$ 


i 




h . 


Ship. 


S 

13 3 

jl 


?!:! 

a-s u; 

Sfll 


o,2 crj 

Ou t) fO 


a 

T3 


II 

ti 




i 


JM!O 


S x 




U 






i) *^ *S U 




d 


ji 




o 


Illl 


i 


. 
jj 


1. 




w 


J 


H 






Greyhound 


/ioi7 
i 845 


10,770 

6,200 


I58-7 


453 


0-350 


Mutine 


/ 977 
\ 757 


9,440 
4,770 


279-5 
109-4 


770 
328 


0-363 
0-334 



3 02 



Theoretical Naval Architecture. 



Resistance. We now have to inquire into the various 
resistances which go to make up the total resistance which a 
ship experiences in being towed through the water. These 
resistances are of three kinds 

1. Resistance due to friction of the water upon the surface 
of the ship. 

2. Resistance due to the formation of eddies. 

3. Resistance due to the formation of waves. 

1. " Frictional resistance? or the resistance due to the 
friction of the water upon the surface of the ship. This is 
similar to the resistance offered to the motion of a train on a 
level line owing to the friction of the rails, although it follows 
different laws. It is evident that this resistance must depend 
largely upon the state of the bottom. A vessel, on becoming 
foul, loses speed very considerably, owing to the greatly 
increased resistance caused. This frictional resistance forms 
a large proportion of the total at low speeds, and forms a 
good proportion at higher speeds. 

2. Resistance due to eddy-making. Take a block of wood, 
and imagine it placed a good distance below the surface of 
a current of water moving at a uniform speed V. Then 
the particles of water will run as approximately indicated 
in Fig. 107 At A we shall have a mass of water in a state of 




FIG. 107. 



violent Agitation, and a much larger mass of water at the rear 
of the block. Such masses of confused water are termed 
"eddies" and sometimes "dead water." If now we imagine 
that the water is at rest, and the block of wood is being towed 



Horse-power, Effective and Indicated, etc. 303 

through the water at a uniform speed V, the same eddies will 
be produced, and the eddying water causes a very considerable 
resistance to the onward motion. Abrupt terminations which 
are likely to cause such eddies should always be avoided in 
vessels where practicable, in order to keep the resistance as 
low as possible. This kind of resistance forms a very small 
proportion of the total in well-formed vessels, but in the older 
vessels with full forms aft and thick stern-posts, it amounted to 
a very considerable item. 

3. Resistance due to the formation of waves. For low 
speeds this form of resistance is not experienced to any 
sensible extent, but for every ship there is a certain speed 
above which the resistance increases more rapidly than would 
be the case if surface friction and eddy-making alone caused 
the resistance. This extra resistance is caused by the forma- 
tion of waves upon the surface of the water. 

We must now deal with these three forms of resistance in 
detail, and indicate as far as possible the laws which govern 
them. 

i . Frictional Resistance. The data we have to work upon 
when considering this form of resistance were obtained by the 
late Mr. Froude. He conducted an extensive series of experi- 
ments on boards of different lengths and various conditions 
of surface towed edgewise through water contained in a tank, 
the speed and resistance being simultaneously recorded. The 
following table represents the resistances in pounds per square 
foot due to various lengths of surface of various qualities when 
moving at a uniform speed of 600 feet per minute, or very nearly 
6 knots in fresh water. There is also given the powers of the 
speed to which the resistances are approximately proportional. 

We can sum up the results of these experiments as follows : 
The resistance due to the friction of the water upon the surface 
depends upon 

(1) The area of the surface. 

(2) The nature of the surface. 

(3) The length of the surface. 

(4) The density of the water. 

and (5) The resistance varies as the ** power of the speed 
where n varies from 1*83 to 2*16. 



304 



Theoretical Naval Architecture. 









L.HNGT 


1 OF Su 


UFACE 


N FEET 








a 






8 


S 


to 




JO 




Si 


it 


si 


i 


|1 


I 


43 


K, 


Nature of surfa ce. 


^'5 


"S . 


;: 


1 


!*'3 


1 


"*'S 


S 




o o 






3 " 







O O 








is 


"Sti. 


SJ 


T3 


y 


~c 


S.J 




.a 


US 


1.5 


2 S 


s, 


B U 


8.S* 


B V 




. G 


I 1 


b C 


It 





II 


*8 


It 




El 

O'S 

* S 


1 
1 


2 
11 


I 


w rt 

If 


.2 
H 


Powe 
resista 


I 


Varnish ... 
Tinfoil ... 


2-00 

2-16 


0-41 
0-30 


I-8 5 
1-99 


0-325 
0-278 


IS 


0-278 
0*262 


1-83 


O-25O 
0-246 


Calico ... 


*'93 


0-87 


1-92 


0*626 


1*89 


0-53I 


I-8 7 


0-474 


Fine sand 


2"OO 


0-81 


2-00 


0-583 


2-00 


0*480 


2-06 


0-405 


Medium sand . 


2-00 


0-90 


2'00 


0-625 


2-00 


0-534 


2*00 


0-488 



And thus we can write for a smooth surface in salt water 



where R = resistance in pounds ; 

S = area of surface in square feet ; 
V = speed in knots relative to still water ; 
/ = a coefficient depending upon the nature and length 

of the surface ; 
w = density of salt water ; 
w = density of fresh water ; 
w/w = 1*025. 

This coefficient /will be the resistance per square foot given in 
the above table, as is at once seen by making S = i square foot, 
V = 6 knots, and w = w . It is very noticeable how the resistance 
per square foot decreases as the length increases. Mr. Froude 
explained this by poirting out that the leading portion of the 
plane must communicate an onward motion to the water which 
rubs against it, and " consequently the portion of the surface 
which succeeds the first will be rubbing, not against stationary 
water, but against water partially moving in its own direction, 
and cannot therefore experience as much resistance from it." 



H or SB-power > Effective and Indicated, etc. 305 

Experiments were not made on boards over 50 feet in 
length. Mr. Froude remarked, in his " report, "It is highly 
desirable to extend these experiments, and the law they eluci- 
date, to greater lengths of surface than 50 feet; but this is the 
greatest length which the experiment-tank and its apparatus 
admit, and I shall endeavour to organize some arrangement by 
which greater lengths may be successfully tried in open water." 

Mr. Froude was never able to complete these experiments 
as he anticipated. It has long been felt that experiments with 
longer boards would be very valuable, so that the results could 
be applied to the case of actual ships. It is probable that in 
the new American experiment tank, 1 which is of much greater 
length than any others at present constructed, experiments with 
planes some hundreds of feet in length may be carried out. 

These experiments show very clearly how important the 
condition of the surface is as affecting resistance. The 
varnished surface may be taken as typical of a surface coated 
with smooth paint, or the surface of a ship sheathed with 
bright copper, the medium sand surface being typical of the 
surface of a vessel sheathed with copper which has become 
foul. If the surface has become fouled with large barnacles, 
the resistance must rise very high. 

In applying the results of these experiments to the case of 
actual ships, it is usual to estimate the area of wetted surface, 
and to take the length of the ship in the direction of motion to 
determine what the coefficient/ shall be. See below for E.H.P. 
due to friction and eddy-making. 

Take the following as an example : 

The wetted surface of a vessel is estimated at 7540 square feet, the 
length being 172 feet. Find the resistance due to surface friction at a 
speed of 12 knots in salt water, assuming a coefficient of 0*25, and that 
the resistance varies (a) as the square of the speed, and (b) as the 1-83 
power of the speed. 

(a) Resistance = 0*25 X I '025 X 7540 X (-) 2 

= 7728 Ibs. 

(b) Resistance = 0*25 X 1*025 x 7540 X (Jf) 1 ' 8 * 

= 6870 Ibs. 2 

1 For a description of this tank, see Engineering, Dec. 14, 1900. See 
discussion on paper by Mr. Baker on Frictional Resistance, I.N.A., 1916. 
2 T.his.hasto be obtained by the aid of logarithms. 



306 Theoretical Naval Architecture. 

It is worth remembering that for a smooth painted surface 
the frictional resistance per square foot of surface is about \ Ib. 
at a speed of 6 knots. 

It is useful, in estimating the wetted surface for use in the 
above formula, to have some method of readily approximating 
to its value. Several methods of doing this have been already 
given in Chapter II., the one known as " Kirk's Analysis " 
having been largely employed. There are also several approxi- 
mate formulae which are reproduced 

(1) Based on Kirk's analysis 

Surface = 2 LD + Y 

(2) Given by Mr. Denny 

Y" 
Surface = ryLD +- 

(3) Given by Mr. Taylor- 

Surface = 1 5 '5 VW.L. 

(4) Used at the Experiment Tank at Haslar 

Surface = '(3 -4 -f ,' 

L being the length of the ship in feet ; 
D being the mean moulded draught ; 
V being the displacement in cubic feet ; 
W being the displacement in tons. 

2. Eddy-making Resistance. We have already seen the 
general character of this form of resistance. It may be 
assumed to vary as the square of the speed, but it will vary 
in amount according to the shape of the ship and the appen- 
dages. Thus a ship with a full stern and thick stern-posts 
will experience this form of resistance to a much greater 
extent than a vessel with a fine stern and with stern-post and 
rudder of moderate thickness. Eddy-making resistance can 
be allowed for by putting on a percentage to the frictional 
resistance. It is possible to reduce eddy-making to a 
minimum by paying careful attention to the appendages and 
endings of a vessel, especially at the stern. Thus shaft 
brackets in twin-screw ships are often made of pear-shaped 



Horse-power, Effective and Indicated, etc. 307 

section, as shown in Fig. 85 E J and Fig. 108. A conical piece is 
always put at the after end of propeller shafts for this reason. 

AFT. FOR* 




FIG. 108. 

The following formula can be used to express the effective 
horse-power due to surface friction and eddy-making in salt 
water, viz. : 

E.H.P. = ^./.S.V 2 ' 88 
V being in knots. 

For the coefficient/, we can take/ = 0-009 f r a length of 
500 feet varying to 0*01 for a length of 40 feet. These values 
are rather greater than would be inferred from Froude's 
experiments, and include an allowance for eddy-making 
resistance. 

On page 332, a table is given for the E.H.P. due to skin 
friction, based on Mr. Froude's constants, assuming the skin 
friction to vary as V 1825 , from speeds of 10 to 40 knots, and 
for lengths of 100 to 1000 feet. The reduction of the co- 
efficient as length increases has been allowed for in this 
table. 

Mr. Baker, in his work on t " Resistance and Propulsion," 
gives the following values of f for salt water and for varying 
lengths in the formula 

Frictional resistance in Ibs. =/. S . V 1825 

where S is wetted surface in square feet 
V is speed in knots. 



Length) 
in feet / 


50 


75 


100 


200 


300 


400 


500 


700 


900 


/ 


0-0096 


0-00935 


0-0092 


0-00898 


0-0089 


0-00883 


0-00877 


0-00868 


0-0086 



See note in " Strength of Shaft Brackets," p. 255, as to resistance of 



308 



Theoretical Naval Architecture. 



These values are obtained by assuming that the frictional 
coefficient of the first 50 feet is the same as that of a 5o-foot 
plank, regardless of the ship's length, and that the remainder 
of the length has the same frictional resistance as the last foot 
of the 5o-foot plank. 

The table given for E.H.P. due to frictional resistance per 
square foot of wetted surface given on page 332 is calculated 
from similar figures to the above, and it is suggested as 
an exercise that some of the figures given be checked. 
Attention is necessary to the units, as the above is for 
resistance in Ibs., and E H.P. = ^ . R . V, so that 

E.H.P. = ^./.V 2 ' 826 

3. Resistance due to the Formation of Waves. A completely 
submerged body moving at any given speed will only experi- 
ence resistance due to surface friction and eddy-making provided 




FIG. 109. 



it is immersed sufficiently ; but with a body moving at the 
surface, such as we have to deal with, the resistance due to 
the formation of waves becomes very important, especially at 
high speeds. This subject is of considerable difficulty, and 
it is not possible to give in this work more than a general 
outline of the principles involved. 



Horse-power , Effective and Indicated, etc. 309 

Consider a body shaped as in Fig. 109 placed a long way 
below the surface in water (regarded as frictionless), and 
suppose the water is made to move past the body with a uniform 
speed V. The particles of water must move past the body in 
certain lines, which are termed stream-lines. These stream- 
lines are straight and parallel before they reach the body, but 
owing to the obstruction caused, the particles of water are 
locally diverted, and follow curved paths instead of straight 
ones. The straight paths are again resumed some distance at 
the rear of the body. We can imagine these stream-lines 
making up the boundaries of a series of stream-tubes, in each 
of which the same particles of water will flow throughout the 
operation. Now, as these streams approach the body they 
broaden, and consequently the particles of water slacken in 
speed. Abreast the body the streams are constricted in area, 
and there is a consequent increase in speed ; and at the rear of 
the body the streams again broaden, with a slackening in speed. 
Now, in water flowing in the way described, any increase in 
speed is accompanied by a decrease in pressure, and conversely 
any decrease in speed is accompanied by an increase in pressure. 
We may therefore say 

(1) There is a broadening of all the streams, and attendant 
decrease of speed and consequent excess of pressure, near both 
ends of the body ; and 

(2) There is a narrowing of the streams, with attendant 
excess of speed and consequent decrease of pressure, along the 
middle of the body. 

This relation between the velocity and pressure is seen in 
the draught of a fire under a chimney when there is a strong wind 
blowing. The excess of the speed of the wind is accompanied 
by a decrease of pressure at the top of the chimney. It 
should be noticed that the variations of velocity and pressure 
must necessarily become less as we go further away from the 
side of the body. A long way off the stream-lines would be 
parallel. The body situated as shown, with the frictionless 
\\ater moving past it, does not experience any resultant force 
tending to move it in the direction of motion. 1 

1 This principle can be demonstrated by the use of advanced mathematics. 
" We may say it is quite evident if the body is symmetrical, that is to say, 



3io Theoretical Naval Architecture. 

Now we have to pass from this hypothetical case to the case 
of a vessel on the surface of the water. In this case the water 
surface is free, and the excess of pressure at the bow and stern 
shows itself by an elevation of the water at the bow and stern, 
and the decrease of pressure along the sides shows itself by a 
depression of the water along the sides. This system is shown 
by the dotted profile of the water surface in Fig. no, which 




FIG. no. 

has been termed the statical wave. The foregoing gives us 
the reason for the wave-crest at the stern of the ship. The 
crest at the bow appears quite a reasonable thing to expect, 
but the crest at the stern is due to the same set of causes. 
This disturbance of level at the bow and stern is described by 
Mr. R. E. Froude as the " forcive " of the actual wave forma- 
tion. If a stone is thrown into water, the sudden disturbance 
propagates a series of waves that radiate in all directions. In 
the case of a ship, the shape of the ship causes the disturbance 
to form diverging and transverse waves as seen below. 

Observation shows that there are two separate and distinct 
series of waves caused by the motion of a ship through the 
water : (i) at the bow, and (2) at the stern. 

Each of these series of waves consists of (i) a series of 
diverging waves, the crests of which slope aft, and (2) a series 
of transverse waves, whose crests are nearly perpendicular to 
the middle line of the ship. 

First, as to the diverging waves at the bow. " The inevi- 
tably widening form of the ship at her entrance throws off on 
each side a local oblique wave of greater or less size according 
to the speed and obtuseness of the wedge, and these waves 
form themselves into a series of diverging crests. These waves 

has both ends alike, for in that case all the fluid action about the after 
body must be the precise counterpart of that about the fore body j all the 
stream-lines, directions, speed of flow, and pressures at every point must be 
symmetrical, as is the body itself, and all the forces must be equal and 
opposite" (see a paper by Mr. R. E. Froude, on "Ship Resistance," read 
before the Greenock Philosophical Society in 1894). 



Horse-power, Effective and Indicated, etc. 3 1 1 

have peculiar properties. They retain their identical size for a 
very great distance, with but little reduction in magnitude. 
But the main point is, that they become at once disassociated 
with the vessel, and after becoming fully formed at the bow, 
they pass clear away into the distant water, and produce no 
further effect on the vessel's resistance." These oblique waves 
are not long in the line of the crest BZ, Fig. in, and the 







waves travel perpendicular to the crest-line with a speed of 
V cos 0, where V is the speed of the ship. As the speed of 
the ship increases the diverging waves become larger, and 
consequently represent a greater amount of resistance. 

Besides these diverging waves, however, " there is produced 
by the motion of the vessel another notable series of waves, 
which carry their crests transversely to her line of motion." It 
is this transverse series of waves that becomes of the greatest 
importance in producing resistance as the speed is pushed 
to values which are high for the ship. These transverse waves 
show themselves along the sides of the ship by the crests and 
troughs, as indicated roughly in Fig. no. The lengths of these 
waves (i.e. the distance from one crest to the other) bears a 
definite relation to the speed of the ship. This relation is that 
the length of the wave varies as the square of the speed at 
which the ship is travelling, and thus as the speed of the ship 
increases the length from crest to crest of the accompanying 
series of transverse waves increases very rapidly. 

The waves produced by the stern of the ship are not of 
such great importance as those formed by the bow, which we 
have been considering. They are, however, similar in character, 
there being an oblique series and a transverse series. 



3 I2 



Theoretical Naval Architecture. 



Interference between the Bow and Stern Transverse Series of 
Waves. In a paper read by the late Mr. Froude at the Insti- 
tution Of Naval Architects in 1877, some very important 
experiments were described, showing how the residuary resist- 
ance 1 varied in a ship which always had the same fore and 
after bodies, but had varying lengths of parallel middle body 
inserted, thus varying the total length. A strange variation in 
the resistance at the same speed, due to the varying lengths of 
parallel middle body was observed. The results were set out 
as roughly shown in Fig. 112, the resistance being set up on a 




340. 



240 i-W 40 

LENGTH OF PARALLEL MIDDLE BODY 
FIG. 112. 

base of length of ship for certain constant speeds. At the low 
speed of 9 knots very little variation was found, and this was 
taken to show that at this speed the residuary resistance was 
caused by the diverging waves only. 

The curves show the following characteristics : 

(1) The spacing or length of undulation appears uniform 
throughout each curve, and this is explained by the fact that 
waves of a given speed have always the same length. 

(2) The spacing is more open in the curves of higher speed, 
the length apparently varying as the square of the speed. This 
is so because the length of the waves are proportionate to the 
square of the speed. 

1 Residuary resistance is the resistance other than frictional. 



Horse-power, Effective and Indicated, etc. 313 

(3) The amplitude or heights of the undulations are 
greater in the curves of higher speeds, and this is so, because 
the waves made by the ship are larger for higher speeds. 

(4) The amplitude in each curve diminishes as the length of 
parallel middle body increases, because the wave system, by 
diffusing transversely, loses its height. 

These variations in residuary resistance for varying lengths 
are attributed to the interference of the bow and stern trans- 
verse series of waves. When the crests of the bow-wave series 
coincide with the crests of the stern-wave series, the residuary 
resistance is at a maximum. When the crests of the bow-wave 
series coincide with the troughs of the stern-wave series, the 
residuary resistance is at a minimum. 

The following formula l gives an approximation to the effec- 
tive horse-power to overcome wave-making resistance, viz. 



The coefficient b, however, has varying values for the same 
ship owing to the interference above mentioned, so that it is 
not a formula that can be relied upon. The total formula for 
E.H.P. can be written 



E.H.P. = /. S . V 2 ' 83 + b . . V B 



where / is a coefficient for surface friction and eddy-making 
appropriate to the length. If 50 per cent, be taken as a 
standard propulsive coefficient at top speed, to 40 per cent. 
at 10 knots, say, values of b can be determined from trial data 
in the user's possession which may be useful for estimating 
purposes. Examples 31 and 32 in Appendix illustrate its use. 

The following extracts from a lecture 2 by Lord Kelvin (Sir 
William Thomson) are of interest as giving the relative in- 
fluence of frictional and wave-making resistance : 

" For a ship A, 300 feet long, 31^ feet beam, and 2634 tons 
displacement, a ship of the ocean mail-steamer type, going at 
13 knots, the skin resistance is 5*8 tons, and the wave resistance 

1 See Mr. Johns' paper, I.N.A., 1907, for a discussion of "approxi- 
mate formulae for determining the resistance of ships," also Prof. Hovgaard, 
1908, I.N.A. 

9 Third volume " Popular Lectures and Addresses," 1887. 



314 Theoretical Naval Architecture. 

is 3*2 tons, making a total of 9 tons. At 14 knots the skin 
resistance is but little increased, viz. 6*6 tons, while the wave 
resistance is 6*15 tons. 

" For a vessel B, 300 feet long, 46-3 feet beam, and 
3626 tons, no parallel middle body, with fine lines swelling out 
gradually, the wave resistance is much more favourable. At 
13 knots the skin resistance is rather more than A, being 
6*95 tons as against 5*8 tons, while the wave resistance is 
only 2 -45 tons as against 3*2 tons. At 14 knots there is a 
very remarkable result in the broader ship with its fine lines, 
all entrance and run, and no parallel middle body. At 14 
knots the skin resistance is 8 tons as against 6*6 tons in A, 
while the wave resistance is only 3-15 tons as against 6-15 
tons in A. 

" For a torpedo boat, 125 feet long and 51 tons displacement, 
at 20 knots the skin resistance was i'2 tons, and the wave resist- 
ance i' i tons." 

Resistance of a Completely Submerged Body. The condi- 
tions in this case are completely different from those which 
have to be considered in the case of a vessel moving on the 
surface. In this latter case waves are produced on the surface, 
as we have seen, but with a completely submerged body this is 
not so, provided the vessel is immersed sufficiently. We get the 
clue to the form of least resistance in the shape of fishes, in which 
the head or forward end is comparatively blunt, while the rear 
portion tapers off very fine. The reason for the small resistances 
of forms of this sort is seen when we consider the paths the particles 
of water follow when flowing past. These paths are termed the 
stream-lines for the particular form. It will be seen that no eddies 
are produced for a fish-shaped form, and, as we saw on p. 306, 
it is the rear end which must be fined off in order to reduce eddy- 
making to a minimum. This was always insisted on very strongly 
by the late Mr. Froude, who said, " It is blunt tails rather than 
blunt noses that cause eddies." A very good illustration of the 
above is seen in the form that is given to the section of shaft 
brackets in twin-screw vessels. Such sections are given in Figs. 
85 E and 1 08. It will be noticed that the forward end is com- 
paratively -olunt, while the after end is fined off to a small radius. 



Horse-power, Effective and Indicated, etc. 3 1 5 

Speed Coefficients. The method which is most largely 
employed for determining the I.H.P. required to drive a vessel 
at a certain speed is by using coefficients obtained from the 
results of trials of existing vessels. They are based upon 
assumptions which should always be carefully borne in mind 
when applying them in actual practice. 

i. Displacement Coefficient. We have seen that for speeds 
at which wave-making resistance is not experienced, the resist- 
ance may be taken as varying 

(a) With the area of wetted surface ; 

(b) Approximately as the square of the speed ; 
so that we may write for the resistance in pounds 

R = K a SV 2 

V being the speed in knots, S the area of wetted surface in 
square feet, and Kj being a coefficient depending on a number 
of conditions which we have already discussed in dealing with 
resistance. 

Now, E.H.P = - , as we have already seen 

(p. 297). Therefore we may say 
E.H.P. = 



where K 2 is another coefficient, which may be readily obtained 
from the previous one. If now we assume that the total I.H.P. 
bears a constant ratio to the E.H.P., or, in other words, the 
propulsive coefficient remains the same, we may write 

I.H.P. = K 3 SV 3 

K 3 being another new coefficient. S, the area of the wetted 
surface, is proportional to the product of the length and girth to 
the water-line ; W, the displacement, is proportional to the pro- 
duct of the length, breadth, and draught. Thus W may be said 
to be proportional to the cube of the linear dimensions, while S 
is proportional to the square of the linear dimensions. Take a 
vessel A, of twice the length, breadth, and draught, of another 
vessel B, with every linear dimension twice that of the corre- 
sponding measurement in B. Then the forms of the two vessels 



316 Theoretical Naval Architecture. 

are precisely similar, and the area of the wetted surface of 
A will be 2 a = 4 times the area of the wetted surface of B, and 
the displacement of A will be 2 3 = 8 times the displacement of 
B. The ratio of the linear dimensions will be the cube root 
of the ratio of the displacements, in the above case ^8=2. 
The ratio of corresponding areas will be the square of the cube 
root of the ratio of the displacements, in the above case 
(4/8) 2 = 4. This may also be written 8*. We may accord- 
ingly say that for similar ships the area of the wetted surface 
will be proportional to the two-thirds power of the displace- 
ment, or W'. We can now write our formula for the indicated 
horse-power 

W* x V 1 
I.H.P. = ILJ^L 

where W = the displacement in tons ; 
V = the speed in knots ; 

C = a coefficient termed the displacement coefficient* 
If a ship is tried on the measured mile at a known displace- 
ment, and the I.H.P. and speed are measured, the value of the 

W* X V s 
coefficient C can be determined, for C = j TT p . It is usual 

to calculate this coefficient for every ship that goes on trial, and 
to record it for future reference, together with all the particulars 
of the ship and the conditions under which she was tried. It 
is a very tedious calculation to work out the term W z , which 
means that the square of the displacement in tons is calculated, 
and the cube root of the result found. It is usual to perform 
the work by the aid of logarithms. A specimen calculation is 
given here : 

The Himalaya on trial displaced 4375 tons, and an I.H.P. 
of 2338 was recorded, giving a speed of 12*93 knots. Find the 
" displacement coefficient " of speed. 

Here we have- W = 4375 

V = 12-93 

I.H.P.= 2338 

1 The coefficients are often termed " Admiralty constants," but it will 
be seen later that they are not at all constant for different speeds of the 
same vessel. 



Horse-power ) Effective and Indicated, etc. 317 

By reference to a table of logarithms, we find 
log 4375 = 3' 6 4io 
log 12-93 = i'in6 
log 2338 = 3-3689 

so that log (4375) 1 = I log 4375 = 2 '4273 
log (i2'93) 3 = 3 log 12-93 = 3'3348 

= 2 ' 4273 + 3 ' 3348 ~ 



= 2-3932 

The number of which this is the logarithm is 247-3, 
accordingly this is the value of the coefficient required. 

2. The other coefficient employed is the " midship-section 
coefficient" 1 If M is the area of the immersed midship section 
in square feet, the value of this coefficient is 

M X V s 
I.H.P. 

This was originally based on the assumption that the 
resistance of the ship might be regarded as due to the forcing 
away of a volume of water whose section is that of the im- 
mersed midship section of the ship. This assumption is not 
compatible with the modern theories of resistance of ships, and 
the formula can only be true in so far as the immersed midship 
section is proportional to the wetted surface. 

In obtaining the W* coefficient, we have assumed that the 
wetted surface of the ships we are comparing will vary as the 
two-thirds power of the displacement ; but this will not be true 
if the ships are not similar in all respects. However, it is 
found that the proportion to the area of the wetted surface is 
much more nearly obtained by using W* than by using the 
area of the immersed midship section. We can easily imagine 
two ships of the same breadth and mean draught and similar 
form of midship section whose displacement and area of wetted 
surface are very different, owing to different lengths and forms. 
We therefore see that, in applying these formulae, we must take 
care that the forms and proportions of the ships are at any rate 
somewhat similar. There is one other point about these 
1 See note on p. 316. 



318 Theoretical Naval Architecture. 

formulae, and that is, that the performances of two ships can 
only be fairly compared at " corresponding speeds." 1 

Summing up the conditions under which these two formulae 
should be employed, we have 

(1) The resistance is proportional to the square of the speed. 

(2) The resistance is proportional to the area of wetted 
surface, and this area is assumed to vary as the two-thirds power 
of the displacement, or as the area of the immersed midship 
section. Consequently, the ships we compare should be of 
somewhat similar type and form. 

(3) The coefficient of performance of . the machinery is 
assumed to be the same. The ships we compare are supposed 
to be fitted with the same type of engine, working with the 
same efficiency. Accordingly we cannot fairly compare a 
screw steamer with a paddle steamer, since the efficiency of 
working may be very different. 

(4) The conditions of the surfaces must be the same in 
the two ships. It is evident that a greater I.H.P. would be 
required for a given speed if the ship's bottom were foul than 
if it had been newly painted, and consequently the coefficient 
would have smaller values. 

(5) Strictly speaking, the coefficients should only be com- 
pared for " corresponding speeds." 2 

With proper care these formulae may be made to give 
valuable assistance in determining power or speed for a new 
design, but they must be carefully used, and their limitations 

thoroughly appreciated. A good method of recording these 

y 

coefficients is to plot them on base of ^. In this way the 

v L 

size of ship is eliminated. 

We have seen that it is only for moderate speeds that th.e 
resistance can be said to be proportional to the square of the 
speed, the resistance varying at a higher power as the speed 
increases. Also that the propulsive coefficient is higher at the 
maximum speed than at the lower speeds. So if we try a 
vessel at various speeds, we cannot expect the speed coefficients 
to remain constant, because the suppositions on which they are 
1 See p. 319. * Seep. 319. 



Horse-power, Effective and Indicated, etc. 3 ] 9 

based are not fulfilled at all speeds. This is found to be the 
case, as is seen by the following particulars of the trials of 
H.M.S. Iris. The displacement being 3290 tons, and the area 
of the immersed midship section being 700 square feet, the 
measured-mile trials gave the following results : 

* I.H.P. Speed in knots. 

7556 18-6 

3958 1575 

1765 12-5 

596 8-3 

The values of the speed coefficients calculated from the 
above are 

Displacement Mid. sec. 

coefficients. coefficients. 

18-6 knots ... 188 ... 595 

1575 , 2l8 .- 690 

12-5 243 ... 770 

8'3 214 ... 677 

It will be noticed that both these coefficients attain their 
maximum values at about 12 knots for this ship, their value 
being less for higher and lower speeds. We may explain this 
by pointing out 

(1) At high speeds, although the "propulsive coefficient" 
is high, yet the resistance varies at a greater rate than the 
square of the speed, and 

(2) At low speeds, although the resistance varies nearly as 
the square of the speed, yet the efficiency of the mechanism is 
mot at its highest value. 

Corresponding Speeds. We have frequently had to use 
the terms " low speeds" and "high speeds" as applied to certain 
ships, but these terms are strictly relative. What would be a 
high speed for one vessel might very well be a low speed for 
another. The first general idea that we have is that the speed 
depends in some way on the length. Fifteen knots would be 
a high speed for a ship 150 feet long, but it would be quite a 
moderate speed for a ship 500 feet long. In trying a model 
of a ship in order to determine its resistance, it is obvious that 
we cannot run the model at the same speed as the ship ; but 
there must be a speed of the model "corresponding" to the 
speed of the ship. The law that we must employ is as follows : 
" In comparing similar ships with one another, or ships with 



320 Theoretical Naval Architecture. 

models, tfo speeds must be proportional to the square root of their 
linear dimensions" Thus, suppose a ship is 300 feet long, and 
has to be driven at a speed of 20 knots ; we make a model of 
this ship which is 6' 3" long. Then the ratio of their linear 
dimensions is 

300 
6^5 = 4 * 

and the speed of the model corresponding to 20 knots of the 
ship is 

20 4- \/48 = 2-88 knots 
Speeds obtained in this way are termed " corresponding speeds" 

Example. A model of a ship of 2000 tons displacement is constructed 
on the \ inch = I foot scale, and is towed at a speed of 3 knots. What 
speed of the ship does this correspond to ? 

Although here the actual dimensions are not given, yet the ratio of the 
linear dimensions is given, viz. I : 48. Therefore the speed of the ship 
corresponding to 3 knots of the model is 

3 \/48 = 20^ knots 
Expressing this law in a formula, we may say 



where V = speed in knots ; 

L = the length in feet ; 
c = a coefficient expressing the ratio V : \/L } and 

consequently giving a measure of the speed. 
We may take the following as average values of the co- 
efficient " c" in full-sized ships : 

When c = 0*5 to 0*65, the ship is being driven at a 

moderate economical speed ; 
c = 07 to 1*0, gives the speed of mail steamers and 

modern battleships ; 

c = i'o to i '3, gives the speed of cruisers. 
Beyond this we cannot go in full-sized vessels, since it is not 
possible to get in enough engine-power. This can, however, be 
done in torpedo-boats and torpedo-boat destroyers, and here we 
have c = 1-9 to 2-5. These may be termed excessive speeds. 

The remarks already made as to wave resistance gives the 
reason for the above. For low speeds the wave-making resist- 
ance is small. When, however, the speed increases such that 
the length of the wave is about the length of ship, we have the 



Horse-power, Effective and Indicated, etc. 321 

maximum interference, and the rate of increase of resistance 
with increase of speed is greatest. If V is speed in knots, the 

V 2 
length of accompanying wave is ; when the wave equals the 

I'o 

V 

length of ship, we have = - 1-33. So that when the ratio 
V L 

V 

-j=- is unity and somewhat above, the resistance is increasing 
v L 

very rapidly. If the speed can be pressed beyond the above, 
we reach a state of things where the wave is longer than the 
length of boat, and although the resistance is very high yet it is 
not increasing at so great a rate. This can only be the case in 
vessels of the destroyer or motor type. The following figures 
show how the total resistance varies in a typical destroyer : 

Up to ii knots as second power nearly, at 1 6 knots as V 3 , 
from 1 8 to 20 knots as (V) 33 , at 22 knots as (V) 2 ' 7 , at 25 knots 
as V 2 , and at 30 knots as V 2 nearly. The maximum rate of 
increase is at 1 8 to 20 knots, and here the accompanying wave 
approximated to the length of the ship. 

Froude's Law of Comparison. This law enables us 
to compare the resistance of a ship with that of her model, or 
the resistances of two ships of different size but of the same 
form. It is as follows 

If the linear dimensions of a vessel be I times the dimensions 
of the model, and the resistance of the latter at speeds V 1} V 2 , V 3 , 
etc., are R 15 Rg, R 3 , etc., then at the ''corresponding speeds" of 
the ship, Vj^T V 2A/Z V 8 V"Z etc., the resistance of the ship will 
be Ri/ 3 , R 2 / 3 , R 3 / 3 , etc. 

In passing from a model to a full-sized ship there is a 
correction to be made, because of the different effect of the 
friction of the water on the longer surface. The law of com- 
parison strictly applies to the resistances other than frictional. 
The law can be used in comparing the resistance of two 
ships of similar form, and is found of great value when model 
experiments are not available. 

In the earlier portion of this chapter we referred to the 
experiments of the Greyhound by the late Mr. Froude. A 
curve of resistance of the ship in pounds on a base of speed 

Y 



322 



Theoretical Naval Architecture. 



is given by A, in Fig. 106. In connection with these experi- 
ments, a model of the Greyhound was made and tried in the 
experimental tank under similar conditions of draught as the 
ship, and between speeds corresponding to those at which the 
ship herself had been towed. The resistance of the model having 
been found at a number of speeds, it was possible to construct 
a curve of resistance on a base of speed as shown by C in 
Fig. 113. The scale of the model was ^ full size,_jmd 
therefore the corresponding speeds of the ship were V "16, or 
four times the speed of the model. If the law of comparison 




125 



SPEED 



FIG. 113. 

held good for the total resistance, the resistance of the ship 
should have been i6 3 = 4096 times the resistance of the model 
at corresponding speeds ; but this was not the case, owing to 
the different effect of surface friction on the long and short 
surfaces. The necessary correction was made as follows. 
The wetted surface of the model was calculated, and by 
employing a coefficient suitable to the length of the model and 
the condition of its surface, the resistance due to surface 
friction was calculated for various speeds as explained (p. 305), 
and a curve drawn through all the spots thus obtained. This 



Horse-power, Effective and Indicated, etc. 323 

is shown by the dotted curve DD in Fig. 113. Thus at 
250 feet per minute the total resistance of the model is given 
by ac, and the resistance due to surface friction by ad. The 
portion 'of the ordinate between the curves CC and DD will 
give at any speed the resistance due to other causes than that 
of surface friction. Thus at 250 feet per minute, these other 
resistances are given by cd. This figure shows very clearly 
how the resistance at low speeds is almost wholly due to 
surface friction, and this forms at high speeds a large propor- 
tion of the total. The wave-making resistance, as we have 
already seen, is the chief cause of the difference between the 
curves CC and DD, which difference becomes greater as the 
speed increases. It is the resistance, other than frictional, to 
which the law of comparison is intended to apply. 

We have in Fig. 106 the curve of resistance, AA, of the 
Greyhound on a base of speed, and in precisely the same way 
as for the model a curve of frictional resistance was drawn in 
for the ship, taking the coefficient proper for the state of the 
surface of the ship and its length. Such a curve is given by 
BB, Fig. 1 06. Then it was found that the ordinates between 
the curves A A and BB, Fig. 106, giving the resistance for the 
ship other than frictional, were in practical agreement with the 
ordinates between the curves CC and DD, Fig. 113, giving 
the resistance of the model other than frictional, allowing for 
the " law of comparison " above stated. That is, at speeds of the 
ship V 1 ^, or four times the speeds of the model, the resistance 
of the ship other than frictional was practically i6 3 , or 4096 
times the resistance of the model. 

These experiments of the Greyhound and her model form 
the first experimental verification of the law of comparison. 
In 1883 some towing trials were made on a torpedo-boat 
by Mr. Yarrow, and a model of the boat was tried at the 
experimental tank belonging to the British Admiralty. In this 
case also there was virtual agreement between the boat and 
the model according to the law of comparison. It is now the 
practice of the British Admiralty and others to have models 
made and run in a tank. The data obtained are of great 
value in determining the power and speed of new designs. 



3 2 4 Theoretical Naval Architecture. 

For further particulars the student is referred to the sources 
of information mentioned at the end of the book. 

Having the resistance of a ship at any given speed, we can 
at once determine the E.H.P. at that speed (see p. 297), and 
then by using a suitable propulsive coefficient, we may deter- 
mine the I.H.P. at that speed. Thus, if at 10 knots the resist- 
ance of a ship is 10,700 Ibs., we can obtain the E.H.P. as 
follows : 



Speed in feet per minute = 10 x 

Work done per minute = 10,700 x (10 x ^p) foot-lbs. 
10700 x 



E.H.P. = 



33000 
328 



and if we assume a propulsive coefficient of 45 per cent. 

X zoo 



45 
= 729 

By the use of the law of comparison, we can pass from one 
ship whose trials have been recorded to another ship of the 
same form, whose I.H.P. at a certain speed is required. It is 
found very useful when data as to I.H.P. and speed of existing 
ships are available. In using the law we make the following 
assumptions, which are all reasonable ones to make. 

(1) The correction for surface friction in passing from one 
ship to another of different length is unnecessary. 

(2) The condition of the surfaces of the two vessels are 
assumed to be the same. 

(3) The efficiency of the machinery, propellers, etc., is 
assumed the same in both cases, so that we can use I.H.P. 
instead of E.H.P. 

The method of using the law will be best illustrated by the 
following example : 

A vessel of 3290 tons has an I.H.P. of 250x3 on trial at 14 knots. What 
would be the probable I.H.P. of a vessel of the same form, but of three 
times the displacement, at the corresponding speed ? 



Horse-power, Effective and Indicated, etc. 325 

The ratio of the displacement = 3 
/. the ratio of the linear dimensions / = -v/3 

= i '44 

.*. the corresponding speed = 14 X v I '44 

= i6'8 knots 

The resistance of the new ship will be /* times that of the original, and 
accordingly the E.H.P., and therefore the I.H.P., will be that of the 
original ship multiplied by / = (1*44)* = 3*6, and 

I.H.P. for new ship = 2500 X 3*6 
9000 

When ships have been run on the measured mile at pro- 
gressive speeds, the information obtained is found to be ex- 
tremely Valuable, since we can draw for the ship thus tried a 
curve of I.H.P. on a base of speed, and thus at intermediate 
speeds we can determine the I.H.P. necessary. The following 
example will show how such a curve is found useful in 
estimating I.H.P. for a new design : 

A vessel of 9000 tons is being designed, and it is desired to obtain a 
speed of 21 knots. A ship of 7390 tons of similar form has been tried, and 
a curve of I.H.P. to a base of speed drawn. At speeds of 10, 14, 18, and 
20 knots the I.H.P. is 1000, 3000, 7500, 11,000 respectively. 

Now, the corresponding speeds of the ships will vary as the square root 
of the ratio of linear dimension /. 

We have 

/*? 

and / = i "07 

. V*"= 1-035 

therefore the corresponding speed of the 739O-ton ship is 
21 -T- i '035 = 20*3 

By drawing in the curve of I.H.P. and continuing it beyond the 20 
knots, we find that the I.H.P. corresponding to a speed of 20*3 knots is 
about 1 1,700. The I.H.P. for the gooo-ton ship at 21 knots is accordingly 

11,700 x / = 11,700 x 1*26 

= 14,750 I.H.P. about 

PROOF OF THE LAW OF COMPARISON. 
Take the following symbols : 
P for force. 
m for mass. 
/ for acceleration. 
/ for time. 
v for velocity. 
/ for length. 



326 Theoretical Naval Architecture. 

Then force P = (mass X acceleration) 

= ( x/) 

velocity = (/ -5- /) 

Acceleration is increase of velocity in unit time, = (/-:- 
Mass varies as the volume or P. 
Force, which equals (in X/), may be written 




/. if -j is constant, force will vary as / 3 . 

Progressive Speed Trials, It is now the usual practice 
to run vessels at a series of speeds from a low speed up to the 
highest speed attainable in order to construct a curve of power, 
etc., on base of speed. Such a record is of the highest value 
as data for design purposes, and the information obtained as to 
slip of propellers will frequently indicate the direction in which 
improvements may be made. At each speed it is necessary to 
obtain simultaneously the revolutions, I.H.P., 1 and speed. The 
usual practice is to run the ship on a measured-mile course. 
Fig. H3A shows such a course. Two pairs of posts, AB and 




| =-.IJ<HfiTL=7 



I SMIPS_ _CouRSg 






Fie. i ISA. 

1 Indicator cards are taken from each piston, showing how the pressure 
of the steam varies at each point of a revolution. A calculation from 
these cards enables the I.H.P. to be determined. For turbine engines no 
corresponding method is available. A method of determining power of 
turbine machinery has been introduced by Mr. Johnson of Messrs. Denny 
Bros., Dumbarton, by measuring the torsion of the shaft by electrical 
instruments. Another method was described by Mr. Gibson of Messrs. 
Cammell Laird (see I.N.A. for 1907). For a general discussion of the 
subject see N.E. Coast Inst., 1908. 



Horse-power, Effective and Indicated, etc. 327 

CD, are placed exactly a knot (6080 feet) apart, and the 
ship's course is steered at right angles. The time of transit 
is taken by a chronometer stop-watch. In order to eliminate 
the effect of tide, several runs are taken both with and against 
the tide, and the " mean of means" is taken. Thus, suppose a 
vessel has four runs, and the speeds observed are 15*13, 14*61, 
15-66, 14-11 knots respectively. Then the "mean of means" 
is obtained as follows : 

First Second Mean of 
Speeds, means means means 
X 2. X 4- X 8. 



I5 ' 13 i 20-741, 

14-61 y/4 }6o'oi 

,,\ 30*27 \ \ 120*05 

*66 6 '-* 



,, 
15*66] o-04 

20*77 
14*113 ** I* 

The true mean speed is therefore 120*05 -r 8 = 15-006 knots. 
The ordinary mean of the speeds is 14-88 knots. The same 
result as the mean of means is got by multiplying by i, 3, 3, i 
and dividing by 8. 

The above is based on the assumption that the speed of tide can be 
expressed as a quadratic function of the time. That is, if y is speed of 
.ide, then 

y = a + a^t -f a^t* 

t being the time, a , a,, a 2 being constants. 

Thus, when t = o, speed of tide y l a t 

t=t 



'=3* J4 = o 
If V is the true speed of ship, then, owing to the tide, the speed at intervals 
of / up and down the mile will be 

(V + j,), (V -^), (V +j 3 ), (V -y.) 
or a mean of means of 



V + 1(3^3 -y*-y*-yi) 

By substituting in the above values for y lt etc., this is seen to be equal to V. 
If six runs are taken up and down, the mean of means is obtained by 
multiplying by I, 5, 10, 10, 5, I and dividing by 32, and it is easily shown 
that if the tide be assumed a cubic function of the time, the " mean of 
means " at equal intervals of time gives the true mean speed. 

It is necessary to run measured-mile trials in deep water, 
or a falling off in speed will be experienced. If the water is 
not deep, the natural stream-lines are not formed round the 
ship, and this restriction is a serious cause of resistance. A 



328 



Theoretical Naval Architecture. 



similar thing is noticed in canals. A conspicuous instance 
was noticed on the trials of H.M.S. Edgar. When tried at 
Stoke's Bay, with a depth of water of 12 fathoms, 13,260 horse- 
power was required for 2oJ knots. On the deep-sea course 
between Plymouth and Falmouth, 2 1 knots was obtained with 
12,550 horse-power, or about f knot difference for the same 
power. In consequence of this, trials at high speeds must be 
carried out on a deep-water course, the finest probably being 
at Skelmorlie, near the Clyde, where the depth of water is 40 
fathoms. 

Colonel English's Experimental Method of deter- 
mining I.H.P. of a New Design by the Use of Models 
(I.M.E., 1896). This method of determining the power for 
a new design is an interesting application of the principles of 
the present chapter. 

Two models are made, one of a known ship, the other of 
the new design, on such scales that when towed at the same 
speed they shall be at the corresponding speeds proper for each. 
In the following table the capitals refer to the ships, and the 
small letters to the models, and the resistance is divided into 
the frictionai and wave-making. It will be remembered that 
the law of comparison only strictly holds for wave-making 
resistance. 





Resistance. 




Sn^^H 




Frictionai. 


Wave-making. 






Actual ship (i) ... 


F, 


w, 


D, 


v. 


New design (2) 


F 2 


W 2 


? 2 


v f 


Model of (I) 


/. 


w l 


4 


v \ 


Model of (2) 


7* 


t 


d, 


Vt 



By the law of comparison 



and if the models are towed at the same speed, v^ z/ 2 ; so 
that 



Horse-power, Effective and Indicated, etc. 329 



4 D 2 A Vi 
This determines the relative scale of models, and 



The total resistance of model (i) = / + w^ and that of model 
(2) =/ 2 + a/ a . Let/ 2 + a> a = (/ + Wj), say. 

The law of comparison indicates that the wave-making 
resistance varies as the displacement, so that 



w, D, 

so that w, = * . 



. + n ./ -/ a 



We want to get the wave-making resistance of the new design, 
viz. W 2 ; we first find w. 2 from the above, and we can calculate 
/[ and/ 2 by the use of appropriate frictional coefficients. To 
get W 1} we proceed as follows : For the known ship we have 
data regarding I.H.P. at speed V tl this can be turned into 
E.H.P. by the use of a propulsive coefficient, and this E.H.P. 



FIG. 1136. 



can be turned at once into resistance, which is (F x + Wj). The 
frictional resistance can be calculated by the ordinary rules, 
and we have left W 1} the wave-making resistance of the known 
ship. The only part of the above expression we do not know 



330 Theoretical Naval Architecture. 

is n. This is obtained by towing the models abreast of one 
another and adjusting so that they are exactly abreast (Fig. 
1133). When this is so, the ratio of the levers determines the 

ratio n. We thus can determine a/ 2 , and W 2 = w 2 . 2 . F 2 can 

a 2 

be calculated, so that F 2 + W 2 is determined. This is turned 
into E.H.P. at the speed V 2 , and, using the same propulsive 
coefficient as before, the I.H.P. is found for the new design 
at speed V 2 . The models were of yellow pine ballasted to 
desired draught. A small electric motor was used for towing, 
and when the levers were adjusted so that the models towed 
abreast, the only measurement necessary was the ratio between 
the levers. 

The method may be made clearer by reference to an 
example. It is desired to know the I.H.P. to drive a destroyer 
of 300 tons displacement at a speed of 30 knots, and a known 
destroyer of 247 tons required 3915 I.H.P. for a speed of 
27*85 knots. The model of this vessel was made on a scale 
of ^, so that the speed corresponding to 27-85 knots was 

--, =6*23 knots. The scale of the model of the new ship 
v 20 

must be such that 6-23 knots of model corresponds to 30 knots 
of the ship, giving a scale of ( - H \ = _L- 

The wetted surface of known ship was calculated to be 
3796 square feet, so that that of model was 3796 x (^j) 2 = 9*5- 
The wetted surface of new ship was 4321 square feet, and of 

(I \ 2 
- ) = 8-02. Using these values and 

appropriate values for the coefficient of friction, we have 

F! (known ship) = 0-0094 x 379 6 X (27'85) 1-83 = 15,720 Ibs. 

/ (its model) = 0-01124 x 9*5 X (6'23) 1>85 = 3-15 Ibs. 

F 2 (new ship) = 0*0094 X 4321 X (3o) 1 ' 83 = 20,500 Ibs. 

/ a (its model) = 0-01124 X 8*02 x (6'23) 1W = 2*66 Ibs. 
The propulsive coefficient being assumed as o'6, we have 
E.H.P. of known ship o'6 X 3915 = 2349, so that the total 
resistance of ship was 



Horse-power, Effective and Indicated, etc. 331 



therefore Wj = 11,747 Ibs. 

From the towing trial at 6*23 knots, n = 0*811, so that 
a' 2 = o'8n. (^) 3 . 11,747 + ' 811 X 3*15 - 2'66 == ro8 Ibs. 
we therefore have 

W 2 = ro8 x (23-2)* = 13,500 Ibs. 

The total resistance of new ship is therefore 34,000 Ibs., and 
assuming the same propulsive coefficient, we have 

I.H.P. = 3^ X 34,000 X 30 X ^ = 5220 

Calculation of E.H.P. Mr. A. W. Johns gave before 
the I.N.A., 1907, a table which gives the E.H.P. due to skin 
friction for a number of speeds and lengths of ship, based on 
Mr. Froude's constants and on the assumption that the skin 
friction varies as V 1825 . 

If S is the wetted surface in square feet, then E.H.P. =/. S, 
where /"has the values given in table, p. 332. 

In the same paper he gave a series of curves based on 
model experiments, reproduced in Fig. 1130, from which, 
knowing the prismatic coefficient of fineness, the residuary 
horse-power can be obtained. The curves are drawn for a 

/ V \ a V 2 
number of values of ( -^\ = (where L is under-water 

length) varying from 05 to 1*3 on a base of prismatic co- 
efficients varying from 0*52 to 0*74. It is very striking to 
note how rapidly the residuary horse-power increases, for high 
values of speed-length ratio, with increase of prismatic co- 
efficient. The prismatic coefficient has been taken with the 
length P.P., and Mr. Johns states that for merchant ships 
better results are obtained by increasing the prismatic coefficient 
by 0*02, this being due to the fact that in such vessels the 
length P.P. is practically the immersed length of the ship, and 
not, as in the majority of warships, an appreciably smaller 
length. In a few ships of exceptionally good form the curves 
give too great a result, but for ordinary forms of ships the 
curves give a good approximation to the results obtained from 



332 



Theoretical Naval Architecture. 



FRICTIONAL RESISTANCE PER SQUARE FOOT OF WETTED SURFACE. 



Speed 
in 
knots. 


Length of ship in feet. 


TOO 


150 


200 


300 


400 


500 


600 


800 


1000 


IO 


0-0188 


0-0186 


0*0184 


0-0183 


0-0181 


0-0180 


0-0179 


0-0176 


0*0174 


II 


0-0246 


0-0243 


0-024I 


0-0239 


0-0237 


0-0235 


0*0233 


0-0231 


0-0228 


12 


0-0315 


0*0312 


0-0309 


0-0307 


0-0304 


0-0302 


0-0300 


0-0296 


0-0293 


'3 


0-0397 


0-0390 


0-0387 


0-0384 


0-0381 


0-0378 


0-0375 


0-0371 


0-0367 


14 


0-0489 


0-0451 


0-0478 


0-0473 


0-0469 


0-0466 


0-0463 


0-0458 


0-0453 


15 


0-0594 


0-0585 


0-0580 


0-0575 


0-0570 


0*0567 


0-0563 


0-0557 


0-0551 


16 


0-0713 


O-O7O2 


0-0697 


0-0690 


0-0685 


0*0680 


0*0675 


0*0668 


0-066 1 


17 


0*0846 


0-0833 


0*0827 


0-0819 


0-0812 


0-0807 


0-0802 


0*0793 


0-0784 


18 


0-0995 


0-0979 


0-0972 


0-0962 


0-0955 


0-0948 


0-0942 


0*0931 


0-0921 


*9 


0-II59 


0*114! 


0-II32 


O-II2I 


O-III2 


0-1105 


0*1098 


o- 1086 


0-1074 


20 


0-1340 


0*1319 


0-1308 


0*I296 


0-1286 


0-1277 


0-1268 


0-1254 


0-1241 


21 


0-1537 


0-1514 


0-I502 


0-I487 


0-I476 


0-1466 


0*1456 


0*1440 


0*1424 


22 


0-1753 


0*1726 


0-1713 


0-1696 


0-I683 


0*1672 


0*1661 


0-1643 


0-1625 


2 3 


0-1988 


0-1957 


0*1942 


0-I923 


0-I908 


0-1895 


0-1882 


0-1861 


0-1841 


24 


0-2242 


0-2207 


0*2190 


0*2l69 


0-2I52 


0*2138 


0-2124 


0-2IOI 


0*2078 


25 


0-2516 


0*2477 


0*2458 


0-2434 


0-24I5 


0-2399 


0-2383 


0-2357 


0*2331 


26 


0-28II 


0*2767 


0*2746 


0-27I9 


0-2698 


0-2680 


0-2662 


0-2633 


0*2604 


27 


0-3126 


0*3078 


0-3054 


0-3025 


0-300I 


0-2981 


0*2961 


0-2929 


0-2897 


28 


0-3466 


0-34I2 


0*3386 


0-3353 


0-3327 


0-3305 


0-3283 


0-3247 


0-3211 


29 


0-3826 


0-3767 


0-3738 


0-3702 


0-3673 


0-3649 


0-3624 


0-3585 


Q-3545 


30 


0-4210 


0-4I45 


0-4113 


0-4073 


0-404! 


0*4014 


0-3988 


0-3944 


0*3900 


31 


0*4624 


0-4552 


0-45 '7 


0-4473 


0-4438 


0*4409 


0-4379 


0-4322 


0-4274 


3 2 


0*5050 


0-4972 


0-4934 


0-4886 


0-4848 


0-4816 


0-4784 


0-4732 


0^4680 


33 


Q'5499 


0-54I4 


0-5372 


0-5320 


0-5278 


0-5243 


0-5208 


0-5I5 1 


0*5094 


34 


0-5995 


0-5902 


0-5857 


0*5800 


0-5755 


0-5707 


0*5679 


0-5617 


o'5555 


35 


0-6508 


0-6407 


0*6358 


0-6296 


0-6247 


0-6206 


0-6164 


0-6097 


0-6030 


36 


0-7047 


0-6938 


0-6885 


0-6818 


0-6765 


0-6720 


0-6675 


0-6603 


0-6530 


37 


0-7611 


0-7494 


0-7436 


0-7364 


o-73>6 


0-7258 


0-7209 


0-7130 


0-7051 


38 


0*8209 


0-8082 


0*8020 


0-7942 


0-7880 


0-7828 


0-7776 


0-7691 


0-7606 


39 


0-8835 


0-8698 


0-8631 


0-8547 


0-8480 


0-8424 


0-8356 


0*8276 


0-8185 


40 


0-9490 


0-9343 


0-9271 


0-9181 


0-9109 


0-9049 


0-8989 


0*8890 


0-8792 



NOTE. The above table has been extended beyond that given in Mr. Johns' 
paper to include lengths of 1000 feet and speeds up to 40 knots. 



Horse-power p , Effective and Indicated, etc. 333 

model experiments. The curves apply to vessels in which the 
ratio beam/draught varies from about 27 to 2*9. For greater 
ratios than the latter the curves give results which are smaller 




than they should be, whilst for smaller ratios than the former 
the results will be too great. 

As an example, take a vessel 500 ft. (P.P.) X 71 ft. X 26ft. 
X 14,100 tons, prismatic coefficient 0-582. Under-water length 
520 ft. 



334 



Theoretical Naval Architecture. 



The approximate wetted surface by Denny's formula 



17 L.D + = i"j X 500 X 26 + 



14,100 x 35 
26 



= 41,100 square feet 
and by Taylor's formula 

J 5'5 Vw.L = 41,200 square feet 

Taking 42,000 and using the coefficients in table, we obtain 
the following values of E.H.P. due to surface friction, from 
16 to 25 knots, viz.: 2860, 3390, 3980, 4620, 5360, 6150, 
7020, 7950, 8960, 10,100. 

Now going to the curves and erecting an ordinate at 
0*582 prismatic coefficient, the values of coefficient at speeds 
16-1, 17*65, 19-1, 20*4, 2 1 '6, 22-8, 23-95, 2 5 knots are 
measured as 0-015, C ' 02I J 0*029, 0*042, 0*060, 0-081, .0-104, 
0*138, which have to be multiplied by (displacement in 




tons) t, giving us 1040, 1450, 2000, 2900, 4150, 5600, 7200, 
9550 residuary horse-power. 

The above results, plotted as in Fig. 1130, give us an 
estimated curve of total E.H.P. 

To obtain the Space which must be passed over 
by a Ship starting from Rest to any Speed short 
of the Full Speed, supposing the Engines are 



Horse-power, Effective and Indicated, etc. 335 

exerting the Thrust corresponding to the Maximum 
Speed. (a) Supposing the resistance is varying as the square 
of the speed. When a ship is being accelerated through the 
water there is a certain amount of water accompanying the 
ship which has to be accelerated as well. This is usually 
taken (based on the Greyhound experiments of Mr. W. Froude) 
as 20 per cent, of the weight of the ship. The virtual mass 

/W\ 
to be accelerated is therefore f . ( J, where g is the accele- 

ration due to gravity (32*2 in foot-second units). 

Let R be the resistance of ship at full speed V. 

r lower speed #. 

Then the force urging the ship is the constant thrust of the 
propeller = R and the force accelerating the ship is R r. 
Now by the principles of dynamics 

Force = mass X acceleration 



or/=f. 



W 



Now / = acceleration = ~ 

__dv ds _ dv 
~ ~ds'~dt~ v "ds 

dv R - r 

so that .=!._,_. or 



W v 



and on integrating- 



w rvi v 

= s ' T } R _ r dv to speed Vj from rest. 



' g 



Now on the assumption that resistance varies as the square 
of the speed 



V ~VV 



336 Theoretical Naval Architecture. 

and on integrating 

W V 2 / V 2 



and the space from speed V x to speed V 2 is 

1 e W Y! i /V 2 - V 2 2 \ 

2 5 - R 10 g< - \ V 2 - V^/ 

() Without making any assumptions as to the variation of 
resistance with speed, if we have a curve of I.H.P. on base of 
speed, we can get a good approximation to the space required 
to go from one speed to another short of the maximum suppos- 
ing the full thrust due to the top H.P. is exerted from the start. 

Take as an example a vessel of 5600 tons, whose I.H.P. at 
speeds of 10, 12, 14, 16, 18 and 20 knots are respectively 950, 
1640, 2720, 4340, 6660 and 10,060. It is desired to obtain 
the space required to increase the speed from 10 to 18 knots, 
supposing the engines are exerting the full thrust corresponding 
to 20 knots. 

Here the virtual weight is x 5600 = 6720 tons, and 
assuming I.H.P. = 2 E.H.P. all through 

101 X 2240 
(I.H.P.),, = 2 X r X v X Q r in tons, v in knots 

(I.H.P.). i ' 






(I.H.P.n 

V J 



_ 

"' R ~ = 137 20 

Taking v as 10, 12, 14, 16, and 18 knots respectively 

i /io,o6o Q'joX 
R - r at TO knots = ~ ~ ~ ~ = 2 9'8 tons 






i /io,o6o i64o\ 
R - r at 12 knots = f ^ --- jy-J = 26'8 tons 

I / 10, 060 2720\ 

R - r at 14 knots = ^(-^- ) = 22< 5 tons 

i /io,o6o 4UO\ , 
R - r at 1 6 knots = -^(-^ --- ^) = l6 ' 9 tons 

i /io,o6o 666o\ 
R - r at 18 knots = ^(-fe --- ^~) = 97 



tons. 



Horse-power, Effective and Indicated, etc. 337 

ds e W v 
Now -=. . 

dv ' g R r 

ds\ 6720 6080 10 

= ' I8 '5 m foot second unlts 



and similarly 

'ds\ , (ds\ ids\ /ds\ 

l = 158,! ) =220,1-77) = 334, ( j =657. 

/ds 
fa. dv, so that we can obtain the integration by 

means of Simpson's first rule having values of -T-. 

We therefore have 
space from 10 to 18 knots 

= \ X 3'38 . [118-5 + 4(i58) + 2(220) + 4(334) + 657] 

= 3600 feet 

(3*38 being the equivalent in foot-second units of 2 knots, the 
interval chosen). 

It will have been seen in the above example that special 
attention is necessary to the units which have been taken as 
tons, feet and seconds. 

The time taken can be obtained in a similar manner by 

integrating values of ( -r ) taken at equal intervals of time. 



Example. The I.H.P. of a vessel of 14,200 tons at 10, 12, 14, 16, 18, 
and 20 knots are 1750, 3150, 5000, 7600, 10,850 and 15,300. Supposing 
the vessel is exerting 15,300 LH.P., how far would the ship travel in 
going from 10 to 18 knots, and how long would it take. 

Ans. 7075 feet. 282 seconds. 

PROPULSION. 

The following notes have been prepared in order to 
provide an introduction to the subject. The subject is too 
large to be dealt with adequately in the space at disposal, and 
for fuller information reference must be made to the systematic 
treatises given at the end of the book. 

Wake. We have dealt above with the various resistances 

z 



338 Theoretical Naval Architecture. 

which oppose a vessel's progress through the water. These 
are mainly frictional and wave making. The friction of the 
water on the surface of the vessel is the cause of a surrounding 
zone of water following in the direction of motion and the 
forward velocity of this zone increases as we go aft. The 
consequence is that at the stern there is a belt of water having 
a forward velocity. This velocity is variable in amount and 
in direction, but may be assumed, in the case of each propeller, 
to have the same effect as a body of water having a certain 
uniform velocity forwards. This body of water is termed the 
frictional wake. The speed of the wake is conveniently 
expressed as a fraction of the speed of the ship, say x . V. 
The wake will have a higher velocity nearer the middle line 
of the ship than at points farther away. The importance of 
this wake is due to the fact that the propeller has thus to work 
in water which has this forward velocity, and therefore the 
speed of the propeller through the water is not the speed of the 
ship V, but (i *)V = V 1} say. 1 The propeller derives 
increased thrust from this cause, and a single screw will benefit 
more than twin screws, owing to the fact above mentioned as 
to the greater velocity of the water nearer the middle line. 
The frictional wake is caused by the motion of the ship, and 
the increased thrust may therefore be regarded as the return 
of a small portion of the energy spent by the ship in over- 
coming the friction of the water on the surface. 

The simple frictional wake above described is complicated 
by the existence of other factors, viz. : 

(a) The stream line wake. (We have seen that the 

stream lines closing round a ship tend to a 
diminution of velocity and an increase of 
pressure.) 

(b) The presence of a wave at the stern. (If the crest 

of a wave is over the propeller the particles 

1 In Froude's notation the speed of wake is expressed as a fraction of 
Vj, say w . V It so that speed of propeller through the water is V w/V^ or 

- t = i +w, w being called the "wake percentage." It follows that 



Horse-power, Effective and Indicated, etc. 339 

of water in their orbital motion are moving 
forwards. If there is a trough the particles are 
moving backwards.) 

The information as to the value of the speed of the wake 
is scanty, and in systematic propeller design it is necessary to 
assume some value. Mr. R. E. Froude assumed 10 per cent, 
of the velocity of the ship as a standard value for the wake, 
i.e. x = 0*1. The following formulae have been obtained as 
the results of Mr. Luke's investigations (I.N.A. 1910) : 

Twin screws x 0-2 -f 0*55 (block coeff.). 
Single screws x = 0-05 4- 0-5 (block coeff.). 

The ratio V x 4- V represents what may be termed the wake 

, V(i - x) 
gam factor, and this is - 5 r= - = i x. 

Augmentation of Resistance. Anything which inter- 
feres with the natural closing in of the stream lines at the 
stern of a ship will cause an increase of resistance. The 
presence of the propeller at the stern is such an interference, 
and gives rise to an augment of resistance. This will be 
greater in a single screw than in a twin screw ship, since the 
propellers in the latter case are further away from the middle 
line of the ship. It is thus seen that, although a single screw 
ship stands to gain more from the frictional wake than a 
twin screw ship, yet it loses more from the augment of 
resistance. 

Thrust Deduction. Instead of regarding this loss as an 
augment of resistance, it is preferable to regard it as a loss in 
the thrust of the propeller. If T be the thrust required to 
overcome the resistance of the ship plus the augment, and R 
the thrust required to overcome the resistance only, then 
T - R is termed the thrust deduction, and T - R = / . T, so 
that R = T(i - /), and (i - t) is called the thrust deduction 
factor. 

Hull Efficiency. The useful work done by the ship is 
the product of the resistance and the speed, or R x V. The 
work done by the propeller is the product of the thrust and the 
speed the propeller passes through the water, or T x V lt 



340 



Theoretical Naval Architecture. 



The ratio (R X V) -^ (T x VJ is termed the hull efficiency, 

and may be written from the above . The usual value 

i x 

assumed for this is unity, the gain due to the wake being 
balanced by the increase of thrust due to the augment of 
resistance. 

The following table, taken from Prof. Dunkerley's recent 
book on " Resistance and Propulsion," expresses admirably 
the way the work available in the engines is expended, and 
what portion of the work is lost beyond recall. The notation 
is that employed in the present notes. 

Work of engines in 

uniform motion 

[I.H.P.]. 



Loss of work 
in friction. 



Work delivered to propeller 
by the shaft [S.H.P.]. 

I 



Available w 
wake = [T( 

We 



me 


ork from Available work due 
V V|)]. to engines 
[P.H.P.J-lTxVJ. 


Loss of work at pro- 
peller due to shock, 
friction, and causing 
useless motions. 


I 
Total available work 
[T.H.P.] = [T X V]. 


rk necessary to Tow-rope work 
vercome aug- [E.H.P.] = [R x V]. 
nted resistance 
[(T - R)V]. 1 


Work available 
in wake. 
1 


Work lost beyond 
recall. 



Horse -Power. We have seen that the horse -power 
necessary to overcome the tow-rope resistance of a ship 
at a given speed is the Effective Horse-Power^ E.H.P. 
Leaving out the constants, E.H.P. = R x V. The thrust 

K H P 
horse-power, U T.H.P. = T . V = - -^-~. The propeller 



Horse-power , Effective and Indicated, etc. 341 

horse-power , P.H.P., makes allowance for the gain due to the 
wake = T X Y! = T . V(i - #), so that 

P.H.P. = T.H.P.(i - x) and P.H.P. = E.H.P. ^-^, 

i.e. E.H.P. = P.H.P. x hull efficiency. 

There are certain losses in the propeller due to the 
frictional and edgewise resistance of the blades and to the 
rotary motion imparted to the water. The ratio between 
the P.H.P. and the shaft horse-power, S.H.P., is the measure 
of the efficiency of the propeller, or P.H.P. 4- S.H.P. = e. 
In reciprocating engines the power actually exerted in the 
cylinders, or I.H.P., is greater than the S.H.P., the relation 
between the two being the measure of the efficiency of the 
machinery, or S.H.P. -r I.H.P. = e m . 

The ratio E.H.P. 4- I.H.P. is the propulsive coefficient, 
and tracing through the various stages, 

E.H.P. _ E.H.P. T.H.P. P.H.P. S.H.P. 
I.H.P. ~ T JLP7 X "PlLP; X S.H.P. X I.H.P. 

= - - X X e X e m . 

i i x 

Taking a case in which the engine efficiency e m = 0*85, 
propeller efficiency e = 0*65, x and / each 0*15, the propulsive 

. 0-85 i 

coefficient is 4 x ^~ X 0*65 x 0-85 = 55^25 per cent, 
i 0*05 

With turbine machinery it is usual to assume that the 
propulsive coefficient is the ratio between the E.H.P. and the 
horse-power being transmitted through the shaft inside the 
ship. Owing to the high revolutions at which turbines work, 
the propellers connected directly to them have a low efficiency, 
and it is found (as e.g. in the Lusitania) that the propulsive 
efficiency thus denned is about 50 per cent. 1 

Cavitation. The force which pushes the ship along is the 
reaction from the projection in a sternward direction of the 
water by the propeller. The momentum of this water, per 
unit of time, is the measure of the thrust which is transmitted 
to the ship through the thrust block. The water will not 

1 The use of the geared turbine enables great all-round efficiency to be 
obtained, as the turbine can run fast and the propeller run slow. 



342 Theoretical Naval Architecture. 

follow up at the back of the blades of the propeller if the 
thrust is too great and if the velocity of the blades is sufficiently 
high. This causes a loss of thrust-producing power, and is 
termed cavitation. Mr. Speakman (Scottish Inst. E. and S., 
1905), from an analysis of numerous trials, considers that to 
avoid cavitation the limit of pressure per square inch of 
projected surface should be about i Ib. for every 1000 feet of 
circumferential velocity of blade tips. Mr. Sidney Barnaby 
assigns nj Ibs. per square inch as the maximum average 
thrust per square inch of projected area. These are for an 
immersion of tip of 12 inches; for each additional foot of 
immersion f Ib. per square inch may be added. This figure 
of nj Ibs., however, may be exceeded for propellers with 
turbine machinery, owing to the uniform turning moment. 1 
The thrust of the screw is obtained as follows : T.H.P. 

P TT p 

= T- 1 -^; / in the absence of definite information may be 

taken as o'i. T.H.P = ^ T X V (V in knots, T in pounds), 

T.H.P. I.H.P. . 

or thrust = 326 = 181 X y , taking a propulsive 

coefficient of 0-5. The I.H.P. is that for the screw in question. 
Taking 11-25 Ibs. per square inch of projected area A p , we 
have 

Projected area of blades 1 _ 181 I.H.P. 

in square feet J ~ A * ~~ 11-25 xT44 X V~ 

I.H.P. 

= O'll X y 

The relation between the developed blade area and 
projected blade area for the Admiralty pattern blade 2 is given 
by Mr. Barnaby as follows : 

Developed area = projected area \/i -f- o'425(pitch ratio) 2 . 

The projected blade area is often expressed as a fraction 
of the disc area, and Mr. Speakman gives the following as 
usual values of this ratio : 

1 Mr. Baker in his book takes 13 Ibs. per square inch to obtain the 
minimum (developed) area. 

2 The Admiralty pattern blade when developed is an ellipse whose 
major axis is the radius of the propeller and whose minor axis is the major 
axis. Propellers for turbines have greater width ratio than this. 



Horse-power, Effective and Indicated, etc. 343 

Reciprocating machinery (naval practice) 
Large ships . , , 0-33 
Destroyers . 0-33 0*4 

Turbines . . . . o - 4 0*56 

The reason for the large value in turbine vessels is the 
excessive speed of rotation, which causes cavitation unless a 
large area is given. The diameter is also brought as small as 
possible in order to avoid excessive velocity of the blade tips, 
which has, however, been as high as 14,850 feet per minute. 

Pitch. A screw propeller usually has the driving face 
(i.e. the after side) in the form of a true screw, and the pitch 
of the screw is denned as the distance this face would advance 
in one revolution if working in a solid nut. If a variable 
pitch is used, then there is an equivalent pitch for the whole 
surface which may be found. The speed of screw is the distance 
it would advance in one minute if working in a solid nut. If 
N is the revolutions per minute and P the pitch in feet, then 
speed of screw is P x N feet per minute. The ratio of pitch 
to diameter (P -f- D) is the pitch ratio /. 

Slip. The advance of a screw through the water when 
propelling a ship is not the speed of the ship V, because of 
the presence of the wake. This speed is V x = (i #)V, where 
^V is the speed of wake. The difference between the speed 
of the screw as defined above and its speed forward relative to 
the water in which it is working is termed the slip, or 



slip = N.P. - y, Vj (Vi in knots) 

N.P.-^.V, 

oo 
The slip ratio, s = ^ p 

This is the true slip, but as we do not generally know the 
value of V 1} the apparent slip is usually dealt with, being the 
difference between the speed of screw and the speed of ship. 

N.P.-^.V 

6o 
Apparent slip ratio s l = - j^p - (V in knots) 



344 Theoretical Naval Architecture. 

Y! being less than V, it follows that the real slip is greater 
than the apparent slip. Cases are on record in which a 
negative apparent slip has been obtained, which means that the 
sternward speed of the water from the screw is less than the 
forward speed of the wake. It is, of course, impossible to 
have the true slip a negative quantity, as this would involve a 
thrust being exerted without the projection of water in a 
sternward direction. 

It follows from the above that the true slip and the apparent 
slip are connected by the following : 

s = x + j x ( i - x). 

Thus, an apparent slip ratio of 20 per cent, with a wake of 
10 per cent, means a true slip of 28 per cent. 

The second volume of Prof. Biles' " Design and Construc- 
tion of Steam Ships" deals exhaustively with the propeller 
question, and gives the methods in vogue for determining 
propeller dimensions based on the model experiments of 
Froude and Taylor. 

See also Mr. Baker's book referred to at the end of the 
book. 

EXAMPLES TO CHAPTER VIII. 

1. The Greyhound was towed at the rate of 845 feet per minute, and 
the horizontal strain on the tow-rope, including an estimate of the air 
resistance of masts and rigging, was 6200 Ibs. Find the effective horse- 
power at that speed. 

Ans. 159 E.H.P. nearly. 

2. A vessel of 5500 tons displacement is being towed at a speed of 
8 knots, and her resistance at that speed is estimated at 18,740 Ibs. What 
horse-power is being transmitted through the tow-rope ? 

Ans. 460. 

3. A steam-yacht has the following particulars given : 

Displacement on trial ... 176*5 tons 

I. H. P. on trial 364 

Speed 13-3 knots 

Find the "displacement coefficient of speed." 

Ans. 203. 

4. A steam-yacht has a displacement of 143*5 tons, and 250 I.H.P. 
is expected on trial. What should the speed in knots be, assuming a 
displacement coefficient of speed of 196 ? 

Ans. I2'2 knots. 



Horse-power^ Effective and Indicated, etc. 345 

5. The Warrior developed 5297 indicated horse-power, with a speed 
of 14*08 knots on a displacement of 9231 tons. Find the displacement 
coefficient of speed. 

Ans. 233. 

6. In a set of progressive speed trials, very different values of the 
' ' displacement coefficient " are obtained at different speeds. Explain the 
reason of this. 

7. Suppose we took a torpedo-boat destroyer of 250 tons displacement 
and 27 knots speed as a model, and designed a vessel of 10,000 tons dis- 
placement of similar form. At what speed of this vessel could we compare 
her resistance with that of the model at 27 knots? 

Ans. 50 knots. 

8. A ship of 5000 tons displacement has to be driven at 21 knots. A 
model of this ship displaces 101 Ibs. At what speed should it be tried ? 

Ans. 3 knots. 

9. A ship of 5000 tons displacement is driven at a speed of 12 knots. 
A ship of 6500 tons of similar form is being designed. At what speed of 
the larger ship can we compare its performance with the 5ooo-ton ship ? 

Ans. 12-53 knots. 

10. A vessel 300 feet long is driven at a speed of 15 knots. At what 
speed must a similar vessel 350 feet long be driven in order that their 
performances may be compared ? 

Ans. i6'2 knots. 

11. A vessel 300 feet long has a displacement on the measured-mile 
trial of 3330 tons, and steams at 14, 18, and 20 knots with 2400, 6000, and 
9000 I.H.P. respectively. What would be the I.H.P. required to drive a 
vessel of similar type, but of double the displacement, at 20 knots ? 

Ans. 13,000 I.H.P. about. 

12. A vessel of 3100 tons displacement is 270 feet long, 42 feet beam, 
and 17 feet draught. Her I.H.P. at speeds of 6, 9, 12, and 15 knots are 
270, 600, 1350, and 3060 respectively. What will be the dimensions of a 
similar vessel of 7000 tons displacement, and what I.H.P. will be required 
to drive this vessel at 18 knots? 

Ans. 354 X 55 X 22-3 ; about 9600 I.H.P. 

13. A vessel of 4470 tons displacement is tried on the measured mile at 
progressive speeds, with the following results : 

Speed. I.H.P. 

8-47 485 

10-43 8Sl 

12-23 J 573 

12-93 2II 7 

A vessel of similar form of 5600 tons displacement is being designed. 
Estimate the I.H.P. required for a speed of 13 knots. 

Ans. 2300 I.H.P. 

14. Verify the figures given for the coefficients of speed of H.M.S. Iris 
on p. 319. 

15. A vessel of 7000 tons requires 10,000 I.H.P. to drive her 20 knots, 
and the I.H.P. at that speed is varying as the fourth power of the speed. 
Find approximately the I.H.P. necessary to drive a similar vessel of 10,000 
tons at a speed of 2iJ knots. 

Ans. 16,000 I.H.P. 



346 



Theoretical Naval Architecture. 



16. Dr. Kirk has given the following rule for finding the indicated 
horse-power of a vessel : 

In ordinary cases, where steamers are formed to suit the speed, the 
I.H.P. per 100 square feet of wetted surface may be found by assuming 
that, at a speed of 10 knots, 5 I.H.P. is required, and that the I.H.P. 
varies as the cube of the speed. 

Show that this can be obtained on the following assumptions : 

/V\ 2 
(i.) The resistance can be expressed by the formula R = f. S. ( -? \ 

where /'= 0*265. 
(ii.) The propulsive coefficient assumed to be about 45 per cent. 

17. Prove that the Admiralty displacement coefficient of speed is the 
same for two similar vessels at corresponding speeds, supposing that the 
efficiency of propulsion is the same. What other assumption is made ? 

1 8. Draw a curve on base of speed of the Admiralty displacement 
coefficient of speed for H.M.S. Drake Qi 14,100 tons, whose curve of I.H.P., 
based on the trial results, give the following figures : 



Speed in knots 


10 


12 


14 


16 


18 


20 


22 


24 


I.H.P. 


195 


3200 


4800 


7000 


10,000 


I 4 ,800 


21,900 


31,000 


\vt x v 


299 


315 


334 


34i 


340 


315 


284 


260 


I.H.P. 



Find also the values of the index n in the formula 

I.H.P.=*.V (,, = log I.H.P.- log I.H.P.,N 

log V, - log V, J 

Ans. (n k) 2-71 (13 k) 2-63 (15 k) 2-83 (17 k) 3-03 (19 k) 373 

(21 )4'ii (23 ) 4-0. 

19. Using the W* coefficient of speed, determine the I.H.P. of a 
vessel similar to the Drake, 555 feet long, at 25 knots, the Drake being 
500 feet long. Ans. 42,600 I.H.P. 

(For further examples, see 28, 30, 31, and 32 in Appendix A.) 

20. A model 20 feet long, wetted surface 77 square feet, has a resistance 
of 26*2 Ibs. in fresh water at a speed of 5 knots. Calculate the effective 
horse-power in sea water of a ship having 16 times the linear dimensions 
of the model and 20 knots speed (/ for model = 0*0104 and for ship 
0*0089, speeds in knots). 

(Durham B.Sc. 1910.) 
The frictional resistance of the model is 

0*0104 x 77 * 5 183 = I 5' 2 lbs - 
The frictional resistance of the ship at 20 knots is 

0*0089 X 77 X i6 2 x 2o 183 = 42,100 Ibs. 

The residuary resistance of the model at 5 knots is 26*2 15*2 = 
ii Ibs., and that of the ship at the corresponding speed of 5/^/16, or 20 
knots, is ii X i6 3 X 1*025 by the law of comparison, allowing lor the 
density of salt water, or 46,200 Ibs. nearly. 



Horse-power, Effective and Indicated, etc. 347 

The total resistance, therefore, of the ship at 20 knots is 42,100 
+ 46,200 = 88,300 Ibs. 

The E.H.P. is therefore 

88,300x^-^5^=5417. 

21. A vessel on successive runs on the measured mile obtains the 
following speeds, viz. : 

27-592, 28-841, 27-965, 28-943, 27-777, 28-426 
knots respectively. 

Obtain : (i) Ordinary average speed, 

(ii) Mean of means of 6 runs, 
(iii) ,, of first 4 runs, 
(iv) ,, ,, of second 4 runs, 
(v) ,, ,, of last 4 runs. 

AHS. (i) 28-257 ; (ii) 28-38 ; (iii) 28-37 ; (iv) 28-418 ; (v) 28-32. 



CHAPTER IX. 
THE ROLLING OF SHIPS. 

NOTE. Throughout this chapter, when an angle is called 
$ or < it is measured in degrees ; when it is called or it is 
measured in units of circular measure, so that 



7T 

[So 



In dealing with the subject of the rolling of ships, it is neces- 
sary to consider first rolling in still water. Although a ship 
will not under ordinary circumstances roll in still water, yet it 
is necessary to study this part of the subject before dealing 
with the more difficult case of the rolling of ships among waves. 

Unresisted Rolling in Still Water. This is a purely 
theoretical consideration because, even if a ship is rolled in 
still water, the rolling will sooner or later cease because of the 
resistances which are set up and which drain the ship of her 
energy. This energy is potential (i.e. due to position) at the 
extremity of each roll, and kinetic (i.e. due to motion) at the 
middle of each roll. At intermediate positions the energy of 
the rolling ship is both potential and kinetic. Work has had 
to be done in the first place to get the ship over, and the ship 
has then stored up in her a definite amount of potential energy. 
This energy is gradually dissipated by the various resistances 
which came into operation until finally the ship comes to rest. 

In a ship rolling we cannot fix upon any definite axis about 
which the oscillation takes place. It appears, however, that 
the centre of oscillation or quiescent point is not far from the 



The Rolling of Ships. 349 

C.G. of the ship, and this point is usually taken as the centre 
of the oscillation. 

The period of oscillation of a ship from side to side rolling 
unresistedly in still water through small angles is given by 



Where m is the metacentric height in feet. 

g is the acceleration due to gravity in foot-second units, 

viz. 32-2. 
k is the transverse radius of gyration of the ship in feet, 

denned as follows : 

(The moment of inertia of a body about any axis is 
found by adding together the product of each 
weight and the square of its distance from the axis. 
If for a ship this axis is through the C.G., W is the 
weight and I the moment of inertia, then k is such 
a quantity that I = W X / 2 and k is the radius 
of gyration. Expressed in mathematical form 
I = W x & = S(w X f).) 

The following is the reasoning leading to the above expres- 
sion for the period, which may be omitted by students not 
having a knowledge of the calculus. 

The equation of motion of the rolling ship is 



where GZ is the righting lever in feet. 

/mass moment N / angular \ _ /couple causing^ 
Y of inertia J x ^acceleration^ ~ ^ the motion J 

5? + S- GZ = 

For ordinary ships the curve of stability for small angles is nearly 
straight, and we can say GZ = m . 0, so that 

d 2 e m . g 

7? + ^- e = ....... < 2 > 

This is a differential equation, of which the general solution is 



(3) 



350 Theoretical Naval Architecture. 

dQ d^Q 
If 6, , are all the same after time /, then 



This is the double oscillation. The period of the single oscillation is 

/ 2 

therefore given by T = ir . / as stated above. 

Equation (3) may now be written 

6 = A . sin - . / + B.cos . / 

and if the initial conditions are assumed, such that when t = o, the ship is 
upright, i.e. 6 = O, and the maximum roll of the ship is 0, this equation 
becomes 

= & sin ^ . / (4) 

or the angle of heel on a time base is a curve of sines. 

The expression for the period is seen to be independent of 
the angle, and it has been found in actual ships that the period 
of roll is practically the same for all angles of oscillation when 
these angles do not exceed 10 to 15 each side of the vertical. 
This is termed isochronous rolling. It is also to be noticed 
that to make the period long, i.e. to increase the time of oscilla- 
tion, it is necessary to 

(i) increase the radius of gyration and, or 
(ii) decrease the metacentric height. 

An application of the above is seen in the current practice of many 
merchant vessels. In some trades, voyages have to be undertaken with 
little or no cargo, owing to the absence of return freights. It is necessary 
for seaworthiness and the proper immersion of the propellers to sink the 
vessel by means of water ballast. This ballast has usually been carried in 
the double-bottom spaces, leading to a low C.G. of the ship, and a large 
metacentric height. The excessive stability causes a short period, and, in 
some cases, has not merely rendered the ship uncomfortable, but actually 
unsafe. The practice, therefore, has grown up of providing spaces for the 
water in other places. Sometimes deep ballast tanks are provided. In one 
patent the triangular space at the side beneath the main deck is made into 
a ballast compartment, and in another the tank top is continued upwards 
to the deck, forming an inner skin at the side, and in the space thus formed 
the water can be carried. It is to be observed that such spaces, exclusively 
devoted to water ballast, are exempt from the measurement for tonnage. 
The added weight of the ballast produces sufficient immersion for seaworthi- 
ness, but does not give excessive stability, and the weight at the sides tends 
to lengthen the period by increasing the radius of gyratiop. 



The Rolling of Ships. 351 

The following are the periods of some typical ships. It 
will be noticed that the heavily armoured battleship of moderate 
GM has a long period, about 8 sec. In the deck protected 
cruisers with no side armour a quicker motion is experienced, 
and in the small classes of war vessel with a relatively large 
GM there is very quick motion. 

H.M.S. Majestic (about 3^ ft. GM and 

great moment of inertia due to side T = 8 sec. 

armour) 
H.M.S. Arrogant, 2nd-class cruiser, deck) _ 

protected \ 

H.M.S. Pelorus, 3rd-class cruiser, deck pro- 1 T _ i 

tected I 

(small period due to (a) small 
Gunboats and mQment Qf inertiaj ^ f T = 2 to 4 sec . 

Destroyers | latively large GM j 

Atlantic liner, small GM T = 10 to 12 sec. 

Passenger yachts T = 5 to 6 sec. 



/ & 

The formula T = TTA / - - fails when the metacentric 
A/ mm g 

height is small. In the particular case of a ship with zero 
metacentric height it gives an infinite period for the roll which 
is absurd. The problem is possible of solution in a wall-sided 
vessel, and this was dealt with by Prof. Scribanti at the I.N.A. 
for 1904. He took three cases, and for each found T m the 
period from his formula and T from the formula above 



m.g 



V 

In a battleship with 3 ft. 1 _T = the errQr being smal] 

GM ) TIB 

T 
In a liner with 4 in. GM . = = 1*31, a considerable error. 



In^almost zero GM, 



j _T = 



352 



Theoretical Naval Architecture. 



For a ship with zero metacentric height, assuming wall- 
sidedness, he found by advanced mathematical analysis * the 
following expression for the period, viz. 



where is the maximum angle in circular measure. 

In the case of a vessel having its curve of stability a curve 
of sines, like a circular vessel or a submarine, the equation of 
motion becomes 

-jp + 2 . sin = o 

This differential equation can be solved by advanced 
mathematical methods, and the following are the periods of 
single oscillations for various angles from the upright, taking 
the period of a small oscillation as unity. 



small 


30 


60 


90 


120 
I'373 


I 5 


180 


I 


i 'oi 7 


1-073 


1-183 


1762 


infinite 



Thus beyond small angles there is an appreciable lengthen- 
ing of the period. The range of stability in this case is 180 
degrees. If for a ship the curve of stability is of the same 
character as a curve of sines, it is reasonable to assume that 
when rolling to angles bearing the same ratio to the range as 
above a similar lengthening of the period would take place in 
comparison with a small oscillation. Thus, for a ship with a 
range of 60 the period for a roll of 20 each side of the vertical 
would be about 7 per cent, greater than that for small angles. 
Although the conditions are not exactly similar to the above 
in the case of ordinary ships the period is lengthened some- 
what for large angles, and this departure from isochronous 
rolling has an important bearing on the safety of a ship when 
being rolled to large angles in a seaway. 

1 Mr. A. W.Johns, R.C.N.C., gave in Engineering, July 18, 1904, a 
method of approximate solution of this problem. 



The Rolling of Ships. 353 

Forces due to Rolling. One application of the equa- 
tion of motion of a rolling ship is to find the maximum force at 

73/1 "2 

any point of a ship when rolling. We can say -jy -f- $ = o, 

72/3 2 

or the angular acceleration = . 6. This is a maximum 

when 6 is a maximum, i.e. at the end of a roll. If we take a 
ship with a period of 5 sec. rolling through 30, 15 each side 
of the vertical, then the angular acceleration at the end of the 

2 

roll is X 15 X = 0-1035 in foot-second units. The 

25 180 

linear acceleration, say 100 feet up, is therefore 10-35 ^ n foot- 
second units. Force = mass x acceleration = X 10-35 

O 

= 0-32 x weight, i.e. a man 100 feet up would have to hold 
on with a force one-third his weight at the end of the roll. 

The following is an example of a similar nature worked 
out: 

A topmast 72 feet in length, height of topmast head being 180 feet above 
water, can be assumed of constant diameter, 15 inches. The ship a/8 seconds 
period is supposed to roll throtigh 30 each side of the vertical. Make an 
estimate of the stress on the material of the topmast at its junction with the 
lower mast, supposing it unsupported by stays. 



In Fig. 114, w is the weight of the topmast, F the transverse force at 
the junction with the mast, L the bending moment, both when at the 
maximum at the extremity of the roll. Then we have 

(a) resolving the forces at right angles to the mast 



w 
- F = - (a + h}^ 



Taking moments about g 



( & is for the topmast and = J 



from which the bend-1 
ing moment 

2 A 




354 



Theoretical Naval Architecture. 



Taking -j- 2 = 7. sin 9, we have 



w is 3520 Ibs., taking the wood of topmast as 4olbs. per cubic foot, a 36, 
h = 108, sin 6 = % 

.'. L = 111,000 ft. Ibs. nearly, 

from which, using - = p the stress at the base of topmast works out to 
4000 Ibs. per sq. in. 




FIG. 114. 

Resisted Rolling in Still Water. It has been found 
by experiment that the rolling of a ship is practically iso- 
chronous, although resistances to the rolling motion are in 
operation. Experimenters on this subject have actually rolled 
ships in order to investigate the laws which govern the motion. 
A small vessel can easily be set rolling by heaving down with 
tackles from a quay. In a large vessel bodies of men can be 
run from side to side, their motion being timed to the ship. 
In the rolling experiments on H.M.S. Revenge (I.N.A. 1895), 
the barbette guns were also trained 15 degrees each side, the 



The Rolling of Ships. 355 

guns being run out first to make the C.G. eccentric. When 
the desired angle of roll was reached, the men and guns were 
stationed at the middle line, while the observations were being 
taken. 

Observing the angles reached on successive rolls a curve 
can be constructed as Fig. 115, the abscissae being numbers of 
rolls and ordinates the angles reached to port and starboard 
successively. Such a curve is termed a a curve of declining 
angles. Fig. 116 shows samples of what is termed a curve of 
extinction, which is obtained from the curve of declining angles, 
the abscissae representing angles of roll and the ordinates 
angles lost per swing. 

It has been found by analyzing a number of these curves 
that the decrement or angle lost per swing can be expressed 
as a<j> + ^$ 2 where $ is the angle of roll in degrees and a and b 
are coefficients which vary for different ships. Thus, calling 
A$ the decrement, we have 

-A$ = 0$ + $ a 
Taking A as a single roll, we have - 

I -g = * + v 

and in the limit this becomes 



which is termed the decremental equation. 
Thus we have 

d& 
Inconstant, T = 8 sec. . . . - = 0-035$ + 0*0051$* 

d<b 
Devastation^ T = 675 sec. . -^ = 0-072$ -f 0*015$* 

, wHhout bilge keels, j _^ = o . oi ^ + 
wUh^bilge keels, j _g = O . o6 ^ + 

The integral form of the decremental equation is 
dn = 




356 



Theoretical Naval Architecture. 




5 10 15 2O 25 30 35 4O 

WtCrriber of rolls 
FIG. 115.- CURVES OP DECLINING ANGLES, Revtnge, 






K> 






I 



47,' 




f 

'^' 



*; 





IO 



15 



ANGLE OF ROLL 

FIG. 1 16 CURVES OF EXTINCTION. Revenge* 



The Rolling of Ships. 357 

which gives the number of rolls to pass from an angle of roll fa to an angle 
<>!. On integrating this becomes 




Thus for a ship in which a = 0-05, b = 0*02, starting from 15 degrees, 13 
rolls are necessary before the angle of roll is 2 degrees. For the Inconstant, 
starting from 15 degrees, the successive angles of roll are 13 '5, 12*2, etc. 

If a ship rolls from 1 port to 2 starboard, supposing the 
curve of stability a straight line, we have 

Dynamical stability at i = J . W . m . ? 
at 2 = i.W.^. 2 2 
so that the loss of energy = JW . m(f - 2 2 ) 

= W . m . m X decrement 
taking J(j + 2 ) = m . 
If R be the moment of the resistance to rolling at angle B, 

,, 

the work done by the resistances from x to 2 is / R . dB. 

2 

We can then equate this work to the loss of energy, since these 
must be equal, viz. 



R . dB = W . m . m x decrement, 
i 

and putting in the value of the decrement from the decremental 
equation above (and remembering that <1> = J 

= W. m. 



( 
J 



This was the method adopted by the late Mr. Froude to 
investigate the laws underlying the resisted rolling of ships. 
i Suppose the moment of resistance varies as the angular 

JJQ 

velocity, or R = k\ . . Then assuming, as in unresisted 

rolling, B = j . sin ^ . /, we have 

dB TT TT 

^=i. T COS T ./. 



358 Theoretical Naval Architecture. 

The work done from to zero is 



7/J 

on putting in the above value for ^ and integrating. Similarly 

for the other side the work done from o to 2 is J . t . ;p. 2 2 . 
So that from x to 2 the work done against the resistance is 
J . ki ^ . m 2 , putting m 2 = (/ + 2 2 ). Equating this to the 

loss of dynamical stability, viz. W . m . TO X decrement, we 
have 

2 

decrement = \ . k^ . ^ rp . OT 

jn 

i.e. a resistance which has a moment proportional to , the 

dt 

angular velocity, will give a decrement proportional to the angle of 
roll. 

2. Suppose the moment of resistance varies as the square 



of the angular velocity, or R = k* . \jjj Then, by a similar 

process to the above and equating the work done by the 
resistance to the loss of dynamical stability, we get that 

decrement = f . k, . w ^ ^ z . m 2 

i.e. a resistance which has a moment proportional to the square 

(dQ\ 
of the angular velocity \^J will give rise to a decrement pro- 

portional to the square of the angle of roll. 

We thus see that if the resistances to the rolling motion 

dQ /d0\ 

are assumed to vary, partly as -r and partly as l~ I , the 

decrement is given by 

2 ^2 



This is of the same form as the decremental equation found 
to fit the curves obtained from rolling experiments. 



The Rolling of Ships. 359 

Mr. Froude attributed the first term to the formation of 
waves, and the second to friction and the passage through the 
water of bilge keels or keel projections (including the flat 
portions of the ship). 

Waves. The rolling motion of a ship creates waves on 
the surface of the water, and these waves pass away and re- 
quire energy for their creation. A wave of very small height 
represents a considerable amount of energy, and the drain on 
the energy of the rolling ship is a distinct resistance tending 
to reduce the rolling motion. 

Friction. This is of small amount, because the surface 
of a ship is kept smoothly painted to reduce the resistance to 
steaming to a minimum. 

Form of Section. If a ship has a sharp bilge, the water 
at the corner has to slip past, and gets a motion opposite to 
that of the ship. The effect both as regards friction and on 
bilge keels is therefore greater than if the section were more 
rounded in form. 

Air Resistance. The resistance of the air to rolling is 
only small under ordinary circumstances, but it may be made 
considerable by the use of steadying sails. If a ship with sails 
set rolls to windward, the wind pressure is increased owing to 
the greater relative velocity, and this the more so the higher 
up we go. The pressure on the sails therefore is greater when 
rolling to windward, and the centre of pressure is higher. 
When rolling to leeward the effective pressure is less, and there 
is a fall in the centre of pressure. 

Bilge Keels. Mr. Froude, in his investigations, took the 
bilge keels as flat surfaces moving through the water, and by 
using data obtained from swinging a flat board in water was 
able to make a calculation for the resistance offered by the 
bilge keels to the motion (see later for the details of this). It 
was found, however, that the observed decrement due to the 
b . < 2 portion of the decremental equation could not thus be 
accounted for. Professor Bryan, in a paper before the I.N.A. 
in 1900, gave further investigations on the subject. Consider 
the flow of water round a right-angled bend as Fig. 117. The 
water adjacent to the surface has to come to rest at the corner 



36o 



Theoretical Naval Architecture. 



and change the direction of its flow. Thus along AB we get 
a diminution of the velocity of the stream lines. With this 
decrease of velocity there must be associated a rise of pressure 
both along AB and BC. Taking now the case of bilge keels 
projecting from the surface as Fig. 118, the ship being sup- 
posed to be rolling clockwise. The relative velocity of the 
ship, and the water along AaAj has to be brought to zero at 
A! and there is caused a rise of pressure along A 2 Ai and 
similarly along A 4 A 3 . These pressures will have resultants as 




FIG. 117. 



FIG. 118. 



P and Q, which with ordinary shaped sections will give a 
moment tending to stop the motion. The effect will be more 
pronounced as the section of the ship is sharper, because of 
the greater relative velocity of the water past the bilge as 
compared with a round section. 

Figs. 115, 116 show very clearly the influence of bilge keels 
in reducing rolling. It was found in the Revenge^ starting in 
each case from 6, that 

without bilge keels 45 to 50 rolls were necessary to reduce to 2 
with bilge keels 8 2 

Curves are also given, showing the effect of motion ahead on 
the rolling. In this case the vessel was proceeding into water 
that was undisturbed by the rolling motion of the ship, and 
the resistance to rolling was somewhat greater than when the 
vessel had no onward motion. 

It has been found that the addition of bilge keels adds 
slightly to the period of rolling, in the case of the Revenge 
about 5 per cent. 



The Rolling of Ships. 361 

The following is the investigation regarding the work done 
by a bilge keel or flat surface, on the assumption that the 
pressure varies as the square of the speed of the bilge keel 
through the water. 

Let A be the area of one side of the bilge keels in sq. feet, 
r the mean distance of the centre of oscillation, 
c the coefficient of normal pressure at i ft. per sec : 
in Ibs. per sq. foot. 



/ JQ\ 2 

Then pressure = c . A . r 2 . ( -j- } at any instant. 

Moment = c . A . r 3 . ( -r. } 

\at / 

(The 2 in the former investigation is therefore c. A . /*.) 

7T 2 

The decrement is given by t*^b*w~~ ^2- m 2 for a 

resistance whose moment is proportional to ( -^ j as seen 
above, or 

decrement = f . c . A . * . 2 . M 2 (W in Ibs.) 



Putting W in tons and T = TT*- we have 

V m .g 

c A . r 3 . P 
decrement 



which is increased, as one would expect, by increase in the 
area of the bilge keels and in the lever. It is also noticed 
that the decrement varies inversely as the I of the vessel, so 
that the bilge keels are proportionally less effective in a vessel 
with large I than with small. The decrement also varies as 
the square of the arc of oscillation, so that when large angles 
are reached, as in a sea way, the influence of bilge keels will 
be most effective. 

Boiling among Waves. A wave is not the passage of 
water, but the passage of motion. The actual movement of 
the particles of water composing a wave is small. The form 
moves with considerable speed, but if a piece of wood be 



362 Theoretical Naval Architecture. 






observed, it is noticed to oscillate about a mean position. In 
the generally accepted trochoidal theory the particles of water 
for deep-sea waves are supposed to move in circular orbits, 
and the diameters of these orbits decrease as the depth 
increases. This orbital motion gives rise to centrifugal force 
and the pressure at the crest of a wave is less than in still 
water, and at the trough the pressure is greater. The buoy- 
ancy, therefore, in the crest portion is less than the normal, 
and in the trough portion it is greater. This is the explana- 
tion of the tenderness of sailing-boats on the crest of a wave. 
The virtual weight is less than the actual, and so the righting 
moment is reduced as compared with still water. The heeling 
moment due to the wind is not affected in this way, and so a 
boat of sufficient stiffness in still water is liable to be blown 
over on a wave. The virtual force of gravity therefore varies 
at different places on a wave, and its direction also varies, 
being perpendicular to the wave profile at any particular point. 
This direction is termed the virtual upright^ and a small raft 
will always tend to place its mast along this virtual upright. 
This has its maximum inclination to the vertical at about a 
quarter the length of the wave from the crest or trough. A 
ship rolling amongst waves will at each instant tend to place 
her masts parallel to a virtual upright, and a surface which is 
normal to each of these virtual upright positions of a ship in a 
wave is termed the effective wave slope^ which is distinctly flatter 
than the actual observed wave profile. 

In dealing with the subject, it is not usual (except for 
sailing-ships) to consider the variation of the amount of the 
virtual weight, but allowance must be made for the variation 
in its direction. Certain assumptions have to be made to 
bring the problem within the scope of mathematical treat- 
ment. These are as follows : 

(a) The ship is lying passively broadside to the wave 
advance. 

(b) The waves are assumed to be a regular series, identical 
in size and speed. 

(c) The waves are assumed long in comparison with the 
size of the ship. 



The Rolling of Ships. 



363 



(d) The profile of the wave is taken as a curve of sines. 

We have first to express the angle the virtual upright makes 
to the upright in terms of the time and other known quantities. 

Fig. i :i 9 represents the construction of the wave, L being 
the | length, H the height (much exaggerated), x and y the 
co-ordinates of a point P referred to axes through the trough. 




FIG. 



119. 



This point P is reached in time /, the time from crest to crest 
being 2T 1} i.e. Tj is the half period of the wave. 

7T L H H 7T 

Then a = ^ . / ; x = ^./; y = - --.cos^./ 



Therefore 



dy TT . H 



where B l is the slope at P. The slope being small, we may 
say that 

_ 7T . H 7T 

*> =' 117 sra T, ' 

which is also the inclination of the virtual upright to the 
vertical. 

From Fig. 120 the equation of motion is, 6 being the angle 
of ship from the vertical 

W // 2 /9 

_^^+W.. (9-00 = 

0_ X being the angle from the virtual upright, 

72/1 2 

or 2 + (0 - 0J = o 



Theoretical Naval Architecture. 






T being the period in still water. Putting in the value of 0, 
found above 



which is Fronde's general equation for unresisted rolling among 
waves. 





FIG. 120. 



Assuming for the initial conditions, i.e. when / = o that the 
ship is upright and at rest in the wave trough, the solution of 
this differential equation is 

T 7T 




i being the maximum wave slope. 

i. Take the special case when T = T : , i.e. the still-water 
period of the ship equals the half period of the wave. This 
is termed synchronism. Putting T = T! in the above equation 
and using the method of the calculus for dealing with inde- 
terminate forms, we have 



, 
= - 



--./cos- 



1 

./J 



When 



= ,, etc. 6 = -^ - y, etc., showing that the 

inclination of the ship is alternately half the maximum wave 
slope. 



The Rolling of Ships. 



365 



When / = o, T x 2T, etc., we have 6 = o, -. 15 TT^ 

^j, etc., *.*. for every half wave that passes an additional 

angle = J . TT x maximum slope, is given to the roll, and thus 
a ship under the given assumptions must inevitably capsize (see 
Fig. 121). Thus the Devastation with a still- water period of 
6f", if lying broadside with no resistance, to waves of J 
maximum slope and period 13^", would increase the roll every 

half wave by -.- degrees, and in 67 J seconds, or rather over 

a minute, would reach 8. Large angles are soon reached 

also, if there is only approximate 

synchronism between the ship and 

the wave. Thus a ship of 5" period 

rolling unresistedly broadside to waves 

of 4" half period with 8 maximum 

slope will, in successive rolls, reach 

11, 20j, 2 7 J. 

These results are borne out by the 
experience of ships at sea. It has 
frequently been observed that ships 
with a great reputation for steadiness 
occasionally roll heavily at sea. This 
is due to the fact that a succession of 
waves has been met with, having a 
period approximately synchronizing 
with the double period of the ship. 
The synchronism may be destroyed 
by altering the course, since what 
affects the ship is the apparent period of the waves. 

2, Suppose the ship has a very quick period as compared with 

T 
that of the wave, so that ^r is small. The equation above 

then reduces to 




FIG. 



i.e. the ship takes up the motion of the wave and behaves 



366 Theoretical Naval Architecture. 

exactly like a raft. The angle of maximum heel will be the 
maximum slope of the wave. 

3. Now take the case in which the period of the ship is 

T 

long compared with that of the wave, i.e. -f^ is small. The 

equation above can be written 

rr-i |~ rr\ * 

= j . '[sin ^ . / - - 1 . sin ^ /J 

T 

This is always small since ^r is small, and the ship will 

never depart far from the vertical. Thus, to secure steadiness 
at sea it is necessary to make the still-water period as long as 
possible. To do this there must be a small metacentric 
height. Such a ship is crank, i.e. easily inclined by external 
forces, but in a sea way is most likely to be steady. 

Atlantic storm waves are about 500 to 600 feet in length, 
and have a period of 10 or n seconds (i.e. 2 . TJ. The 
longest wave recorded had a length of about 2600 feet, and 
a period of 23 seconds. The battleship and liner, quoted 
above as having periods of 8 and 10-12 seconds respectively, 
should therefore prove steady ships in a sea way, as synchro- 
nism would only be experienced when meeting with waves of 
periods 1 6 to 24 seconds, which are quite exceptional. 

Resisted Rolling among Waves. If we take the 
critical case of a vessel meeting with waves whose half period 
is equal to the ordinary period of the ship, then the angle for 

7T 

each swing is increased by - . $ a as seen above. The decre- 
ment due to resistance is given by 0< + b<$ and the increment 
per swing is therefore . $1 (a*j> + & 2 ). The angle of swing 
will go on increasing until an angle of roll is reached such 

7T 

that - . 3>j = a& + ^ 2 . The increase due to synchronism is 

then just balanced by the decrement due to resistance, and we 
get a steady roll of 3>. We have seen above that when large 
angles are reached a ship is not isochronous in her rolling, 
and also that the fitting of bilge keels causes an increase in 



The Rolling of Ships. 36; 

the period. Therefore, under the actual conditions obtaining, 
with a synchronous swell the ship will not necessarily capsize, for 

(a) As large angles are reached the ship departs from 

isochronous rolling. 

(b) Resistances come into operation, and there is also 

the further condition, viz. : 

(c) A succession of waves of precisely the same period 

is a very unlikely occurrence. 

Apparent Period of Waves. We have spoken above 
about the apparent period of waves as affecting a ship's 
rolling. If ft is the angle the direction of the ship's advance 
makes with the crest line of the wave, then if v be the speed 
. of the ship, v . sin ft is the speed of the ship against the wave 
advance \ and z/ being the speed of the wave, the waves will 
meet the ship at a speed z> + v . sin ft. If T be the actual 
period of wave, then the apparent period T' is given by 

T = TO 4- (i + ^ sin ft). If ft is negative, ft' say, i.e. the ship 

travels away from the wave advance, T' = T -f- (i . sin/3'). 

Thus in the first case the apparent period is diminished and in 
the second case increased. 

Graphic Integration of the Rolling Equation. 
i. Unresisted Rolling in Still Water. The equation of motion 
of a ship rolling unresistedly in still water has been seen to be 

^0 g 

^ + ^-GZ = o 

This cannot be mathematically integrated, because there is 
no relation between GZ, the righting arm and the time. By 
assuming that GZ = m . 6, a. solution can be found leading to 

/ A3 

the expression, T = TTA / , being small. This enables 

V m.g 

the equation of motion to be written 

^0,^ GZ_ 
<// 2 + T 2 ' m = 

where T is the time of oscillation from side to side, or one 
half the mathematical period. 



368 



Theoretical Naval Architecture. 



By assuming GZ = m . sin 6 we have the case of a circular 
ship or a submarine, and the equation can be solved by 
advanced mathematical methods, the solution for various angles 
being given on p. 352. 

By the process known as " graphic integration " the solution 
can be accurately found and the process can be extended to 
the case when resistances operate. 




FIG. 122. 



To lead up to the subject, take the case of a body falling 
freely under gravity. The force causing the motion is constant, 
viz. that due to gravity = P, say. Now force = mass x accele- 



ration, or P = /i and/= ^-, /. v f P . dt taking unit mass. 

b 



dt 



If therefore we have a force curve on base of time, Fig. 122 
(in this case a straight line), the velocity is found by integrating 

the force curve. Again v = -r. or s = jv dt, i.e. the space or 

position is found by integrating the velocity curve. Thus, in 
the figure the body would have fallen the distance 4D in time 






The Rolling of Ships. 369 

*, = 4 seconds. EH being the force curve, OBK the velocity 
curve, and OBD the position curve. 

Now take the reverse process. Having given the position 
curve OD, at any point, as D, the tangent makes an angle with 

the base 6 such that tan 9 = -^ , i.e. the velocity. At times 

o, i, 2, 3, 4, etc., seconds, the velocities are o, g, 2g, 3^, 4^, etc., 
i.e. tan 6 has values o, *, 2g, etc. Setting down as in lower 
figure, the tangents to the position curve at A, B, C, D, etc., 
are parallel to Oa, O, O, O</, etc. 

The position curve is the second integral from the force 
curve, and conversely the force curve is the second derived 
from the position curve, and the intersection of tangents at two 
points of the position curve is below the centre of gravity of the 
corresponding portion of its second derived ', i.e. the force curve. 
(For proof of this see later.) 

Thus to get A we find the C.G. of OE, and at T draw TAV 
parallel to Oa in the lower figure. For the second interval we 
draw OAW parallel to O^, and so on with succeeding intervals, 
which enables the position curve OABCD to be drawn in. 

The equation we have to solve is 

J^? = 7r! GZ 

df ~T 2 ' m 
dropping the sign, 

dm TT GZ 

^ = T~ 2 '^r 

dO . 
to = , the angular velocity in circular measure. 

In a small time A/ the change of angular velocity is therefore 

"* GZ A, 

Aw = j A/ 

T^ m 

It is more convenient to use degrees, so that 

7T 2 180 GZ 

Aw = A/ degrees, 

1 TT m 

and taking the interval of time . T, 
1 80 GZ 



2 B 



370 



Theoretical Naval Architecture. 



If therefore at a certain interval the angular velocity is 
represented by the slope of the line OB = tan a, Fig. 123, on a 
base AO = i'oi3 T, and BD is set up equal to the mean 

value of over the succeeding interval ^j T, then the 




FIG. 123. 



FIG. 124. 



slope of OD given by tan /? is the angular velocity at the 
end of the interval -^ T, for 

/i8o GZ\ 

AK H- &IJ . \ TT ' tfl 

tan 



AO 



BD V 

= tan a + 



m / 



I-OI3 1 



If the force curve in this case is one where ordinates are 

' and if force and position curves are as shown in 
TT m 

Fig. 124 on a time base, then at any ordinate at time / the 

180 GZ 
position must correspond to the value of at that 

angle. Where these two curves intersect on the base line, as 
they must do simultaneously, we have the value of the half 
time of the oscillation supposing we start from : the initial 

angle. At the angle OA there is a definite value for ^- - 

= OB, and we now by a process of trial and error have to 
find the curves AP and BP. In the first place we draw a 






The Rolling of Ships. 



371 



modified curve of stability on angle base whose ordinates are 
values of (the slope of which is 45 to the base line), as 

Fig. 125. A convenient scale to use is found to be J" = i, 
J" = i unit of force, and i" = ^ T. 

Set off equal intervals of time ^ T on the base line, i, 2, etc. 



2O 4-O 6O 8O IOO 
MODIFIED STABILITY 
CURVE 

FIG. 125. 




FIG. i5A. Graphic integration for a simple pendulum. 

(Fig. 1 25 A), and mark off OA equal to the initial angle assumed. 
Then at angle O A the modified curve of stability has the value 

T 

which is set down OB. Now over the first interval the 

10 

force will vary (diminishing if we start before the angle of 
maximum stability and increasing if we start after this angle). 



372 Theoretical Naval Architecture. 

An assumed slope BD is taken for the force curve. The mean 
value over the interval is LK ; the slope given by LK -i- 1-013 T 

T 

will give the change of angular velocity in interval . Draw 

AC parallel to base line and the position curve must be a 
tangent to AC at A, since velocity at starting from the extreme 
angle is zero. The e.g. of the force curve OD is then found 
and squared up to meet AC in C. From C draw CGF parallel 

LK 

to the slope given by tan a = . Now we check to see 

1*0131 

if at the angle iG the ordinate of the modified curve of stability 
is iD. If not, the process must be repeated until an agree- 
ment is found. We then proceed to the second interval and 
guess in DE and find F over the e.g. of lE. FH is drawn 

MN 
parallel to the slope given by tan fi = tan a -f ^ 2H and 

2E are again checked as before. In this way, by a process of 
trial and error, the position curve and the force curve may be 
obtained and faired in. They must meet simultaneously on 
the base line. 

Fig. 1 25 A shows the diagram worked out for the case of a 
simple pendulum, or a submarine or circular vessel, for which 
the equation of motion is 

#0 , 7T 2 

^a + ,p sm = o 
and the ordinate of the modified curve of stability is sin 6. 

TT 

The initial angle OA is taken as 120, and it is seen that the 
curves cross the base line together at an abscissa of o'685T and 
double this, viz. i*37T, is the period of the single swing from 
120, T being the time of a small oscillation. The working 
of this problem for various initial angles is recommended as 
an interesting exercise, the results for 30, 60, 90, 120, 
150 should be I'oiy, 1*073, I ' I ^3) J '373i 1762 times T 
respectively. 

Example. H.M.S. Devastation, with GM of 4 feet, has a curve of 
stability whose ordinates every 5 are o, 0-36, 0*69, O'8o, 0*82, 078, O'66, 



The Rolling of Ships. 



373 



0*44, O'2O, and range 43?. The period T for small angles is 675 sec. 
Find the period if heeled to (a) 20, (l>) 30 from the upright and allowed 
to roll freely. 

Ans. (a) 8-1 sec., (b) io'3 sec. 

In the above we have to find graphically the C.G. of a 
trapezoid with reference to an ordinate. This is found as 
follows : Make AC = J . AE (Fig. 126), DQ being the middle 
ordinate. Join CQ and draw SH parallel to the base line, 
then H is in the same abscissa as the C.G. of the trapezoid. 

Proof. The shift of G. from middle ordinate is due to the shift of the 

triangle QS to the position aRQ through a distance of if . h, 2h being the 

base. The area of triangle QS is . h . PQ. The moment of transference 

is therefore 3 . PQ . A 2 . This equals (area of rectangle) X (shift of C.G. x). 

J 



or 



h - 
-- 



This may also be proved from the result of Example 22, Chap. II., which 
is left as an exercise. 





FIG. 126. 



FIG. 127. 



Proof that the intersection of tangents at two points of a curve 
is at the abscissa of the centre of gravity of the corresponding 
portion of its second derived curve. Let the curves EFG, OCD, 
OAB (Fig. 127) be three curves such that any ordinate of OCD 
is equal to the integral of EFG up to that point, and any 
ordinate of OAB is equal to the integral of OCD up to that 



374 Theoretical Naval Architecture. 

point. Then the tangents to the curve OAB at the points A and 
B will intersect at the abscissa of the centre of gravity of the 
area HG. 

EFG is the second derived from OAB. 
Let equation of EFG bejy =/(#) 
OCD be ;>=/'(*) 
OAB be y =f'(x) 
Then we have /' (*) = jf(x)dx, and/"(*) = Jf(x)dx 

or/w = (x} and/ ' (x) = f " (x) 



Take two abscissae x 1 and x 2 for which the ordinates of EFG 
nd^ 2> of OCD are ^/ and j' 2 ' and of OABj^/' and y* . 



The abscissa x of the C.G. of HG is given by 



__ 



r(f\. 

J *i\dx} 



Now the tangent to OAB at A has the equation 

> -*-(&?-** 

and the tangent at B has the equation 



The Rolling of Ships. 375 

Solving for x will give their intersection, or 



).- <).+*- 







dx/ z >!/*/, 

which is the same expression as the abscissa of the centre of 
gravity of HG found above. 

We now have to consider the case of : 2. Resisted Rolling in 
Still Water. We have seen that with resisted rolling the decre- 
ment on a single swing can be written 

in circular 



2 <?. 

or ~ ' = 4 --- 



i.e. a and in the decremental equation d$ = a$ -f- ^^ 2 can 
be written 



, __ 4 7T 

" 5 ' 2 ' 



W . m . T 2 
from which ^ and k z can be determined if a and b are known. 

7 n //7fl\% 

The moment of resistance = & . + / 2 . ( ) 

a / \ a / / 

Substituting for the unknowns ki and / 2 we have 
Moment of resistance 

n circular 
measure 

. T d& . , T 2 , /^d> \ 2 1 d<b C in de- 



The equation of motion of the rolling ship is now 

W 2 d*Q 

g ' ' dt z 



+ W . GZ = o 



376 Theoretical Naval Architecture. 

d z e 7T 2 



and for in degrees we have 



= o 
J 

In the former investigation we multiply GZ by , and we 

do the same for the resistance, which becomes 
2 T d<j> T 2 

~^' a '~dt + *v j 

which is termed the "resistance indicator" This has to be 
added to the modified force, if the ship is swinging away from 
the upright when resistance acts with stability in stopping the 
motion, and subtracted if the ship is swinging towards the 
upright when resistance acts against the stability. 

If -? *= 10 degrees per second, then the ordinate of the 
eft 

resistance indicator is 

2 T 5 T 2 . 

__,a. IO + f ._.. IOO 

This would be set off as an ordinate at a point C (Fig. 128), 
such that AC -J- 1*013 T = 10 degrees per secondhand so on. 

Guess in CF resisted and CG unresisted. AD is the mean 
force over the interval, and we set up OL = AD. Then the 
slope of LQ gives the angular velocity at the end of interval, 
MP is therefore the slope of position curve. Then the angle 
BN should give on the modified stability curve the distance 
BG, FG being equal to the ordinate of resistance indicator at 
L. Thus by a process of trial and error we obtain a series 
of tangents to the position curve and this crosses the base line 
simultaneously with the stability curve. Proceeding past the 
upright we should obtain not only the time of the single 
oscillation but also the angle from the upright to which the 
vessel rolls. The values of a and b for the resistance indicator 
are obtained from the decremental equation for the ship. 



The Rolling of Ships. 377 

As exercises, the " Inconstant " may be taken whose decre- 
mental equation gives a= 0-035, b = 0*0051. Starting from 30 




FIG. 128. 



I-O/3J 



the angles reached in successive rolls are 25-2, 2 1 -6. Starting 
from 25 the angles reached are 21-3, 187. T = 8 sec., 



378 



Theoretical Naval Architecture. 



GM = 2*3 feet. Stability curve at 10 intervals o, 0*5, 1*03, 

i*7> 2 '43. 2 '75> 2 ' 6l > 2 * 6 > i'55 

3. Unrests ted Rolling among Waves. For a ship broadside 
on to a given wave the stability at any instant is determined 
by the angle between the centre line of the ship and the virtual 
upright, i.e. the normal to the wave surface. This is the angle 
between Oa and Ob in Fig. 129. We therefore draw on our 




FIG. 129. 



base line a curve of wave slope which is taken as a curve of 
sines 



Thus for a wave of maximum slope 8 and 8" half period we 

This then is the base line from 



. 180 
have i = 8 X sin -5- 

o 



which to measure the angle of inclination to the virtual upright. 
The process is then carried on as before. In Fig. 129 is 
worked out the case of a ship upright and at rest in the wave 
trough with straight line stability on a wave of 80 maximum 
slope, and whose half period T x is the same as the time of 
oscillation T of the ship. These conditions are known from 
previous investigations to lead to an increment of roll to every 

half wave of - X 8 = 12^. The tangents to the position curve 



The Rolling of Ships. 379 

are parallel to the lines drawn from i, 2, 3, etc., to the point 
given by i'oi3T. The force curve is checked interval by 
interval with the angle from the virtual upright, and must cross 
the base line at the abscissa of the intersection of the wave 
slope curve with position curve. If the example in Fig. 129 
be continued to 20 intervals or over the complete wave, an 
angle of 25 would be reached. This is recommended as an 
exercise. The process can, of course, be applied for a given 
curve of stability and any assumed conditions for period of 
ship, period of wave, and maximum slope of wave. 

4. Resisted Rolling among Waves. In this case the process 
is similar, only the effect of the " resistance indicator " is 
brought in as in 2. 

For an exhaustive account of the application of the process 
of graphic integration, see Sir W. H. White's paper (I.N.A. 
1 88 1 ) on the rolling of sailing ships. In this paper, in addition 
to resisted rolling among waves, account was taken of (a) 
moment due to pressure of the wind on the sails, and (b) the 
variation of the virtual weight in different portions of the wave, 
this being necessary as this variation affects the righting 
moment, while (a) is not thus affected. 

Pitching. The expression for the period of pitching of 
a ship is of a similar form to that for rolling, but we have to 
use /&! the radius of gyration of the vessel about a transverse 
axis through the centre of gravity of the vessel, and GM t the 
longitudinal metacentric height. This period for a single 
oscillation is therefore 



It would be desirable, if other conditions allowed, to make 
the period of pitching as small as possible, and ships with the 
heavy weights concentrated near midships are found to be 
better sea boats than vessels with heavy weights at the ends. 

EXAMPLES. 

I. A vessel of 13,500 tons displacement has a GM of 3^ feet and a 
period of 8 seconds. Find the period of roll when 600 tons of coal are 
added each side of the vessel in a bunker 21 feet deep and 9 feet wide, 
the C.G. of the bunkers being II feet below the original C.G. of the 



380 Theoretical Naval Architecture. 

ship, and 26 feet out from the middle line. The vessel has a horizontal 
curve of metacentres over the limits of draughts corresponding to the above 
conditions. 

V& 
from which & = 823, so that 
m 

I = 13,500 x 823 = 11,100,000. 

The addition of the coal n'down pulls down the C.G. of the ship 

I 2OO X II 

'- - = o'9 ft., making the GM 4*4 ft. 

We now have to calculate the new I about the new C.G., I of coal 
about old C.G. is given by 

2[( T ' 5 X 600 X 9 2 + 600 X 20 7 ) + ( T ^j X 600 X 21* + 600 x ~n-)] = 1,008,600 
I of total about old G.C. is accordingly 

11,100,000+ 1,008,600= 12,108,600 
and about the new C.G. 

12,108,600 (14,700 x o'9 2 ) = 12,096,700 
The new k 1 is therefore 

^96,700 

14,700 
and the new period is accordingly 



2. A cruiser of 5000 tons has a metacentric height of 2'8 feet, a period 
of 7 seconds, and a horizontal curve of metacentres. Calculate the period 
when two fighting tops of 10 ions each are added to the ship at a height 
of 70 feet above the C.G. 

Ans. 7 '5 sees. 



CHAPTER X. 
THE TURNING OF SHIPS. 

WHEN a ship is moving ahead and the rudder is placed 
obliquely to the middle line, the streams of water which flow 
aft relative to the ship are deflected in their course and 
give rise to a resultant pressure normal to the plane of the 
rudder, as P in Fig. 130. The calculation of the amount of 
this normal pressure will be dealt with later, but it may be 
stated here that it depends on 
(i) The area of the rudder, 
(ii) The shape of the rudder. 

(iii) The angle at which the rudder is placed to the centre 
line. 

(iv) The square of the speed of the water past the rudder. 

The area of the rudder is usually expressed in terms of 
the area of the longitudinal middle line plane of the ship, or 
approximately the length times the mean draught. 

or area of rudder = L.D. 

m 

The value of m varies considerably. In large war vessels 
it is from 40 to 50, but in exceptional cases, where great 
manoeuvring power was desired, it came out to 33. In the 
Lusitania its value was about 60. 

As regards the shape, the pressure for a given area will be 
appreciably greater for a narrow, deep rudder than for a broad, 
shallow rudder. 

The usual maximum angle to which rudders are put is 35 
to the centre line. 

In a sailing vessel the speed of the water past the rudder 
is rather less than the speed of the ship, because there is the 
frictional wake. The friction of the water on the surface of 
the vessel induces a current of water in the direction of the 



382 Theoretical Naval Architecture. 

ship's motion so that at the stern the water a short distance 
away from the ship has a forward motion. In a sailing ship, 
in order to get pressure on the rudder, it is necessary that the 
ship shall be in motion, and such a ship loses her power of 
steering as she loses way. 

In a ship driven by a propeller, although there is the same 
frictional wake, the action of the propeller sends a stream of 
water astern, so that such a ship has steerage directly the 
engines are working, a very great advantage. And this effect 
will be greater for a ship having the propeller in line with the 
rudder, as in a single-screw ship and in ships with double 
rudders like the Dreadnought^ than in ordinary twin-screw 
vessels. Although in screw vessels there is the frictional wake 
mentioned above, the speed of the water past the rudder will 
be appreciably greater than the speed of the ship, because the 
speed at which the water leaves the propeller is greater than 
the speed of the ship, the difference being known as the slip. 

In any case it is absolutely necessary for good steering 
that the water shall get a clean run past the rudder. Vessels 
with very full sterns have been found to steer very badly. 

In a ship having a deadwood in front of the rudder the 
slackening of the speed of the streams of water gives rise to a 
side pressure which has a considerable influence in pushing the 
ship over at the start. This is specially noticeable in boats. 
But for good turning the deadwood is unfavourable, as will be 
seen later, and in ships designed for great manoeuvring power 
the deadwood is always cut away. See sterns on pages 

390, 39i- 

In Fig. 130 let P be the normal pressure acting on the 
rudder at C. Introduce at G, the centre of gravity of the ship, 
two equal and opposite forces of value P parallel to the line 
of action of P. 

Then we have acting on the ship 

(i) A couple tending to give angular motion of amount 

P X DG ; and . 
(ii) a force P acting in the line EG. 

The couple is approximately equal to P X X cos 6, and 



The Turning of Ships. 



383 



if P is taken to vary as sin 0, the couple will vary as 
sin 6 X cos or sin 26, and it can be shown that this is a 
maximum when 6 = 45. The sine law, however, for the 
pressure is known to be incorrect, arid it is very probable that 
the usual practice of 35 as the maximum helm angle gives 
the maximum turning effect. Indeed, it appears quite possible 
for certain shapes of rudders that angles of helm as low as 25 
will give as good turning results as 35. 

The force P acting in the line EG may be resolved into its 
components. 

(i) FG, tending to move the ship bodily sideways. The 
motion in this direction, however, is small because of 
the great resistanceof the ship to the side motion. 




FIG. 130. 

(ii) EF, in a fore and aft direction, and this has a sensible 
effect in checking the speed of the ship. 

Path of Ship when Turning. On putting the rudder 
over the ship will commence to turn in a spiral path, of which 
several examples are given later, and this path soon becomes 
approximately circular. The distance from the position when 
the helm is put over to the position when the ship is at right 
angles to her original course is termed the advance. The 
distance from the position when the helm is put over to the 
position when she has turned through 180 or 16 points is 
termed the tactical diameter. 

In a ship thus turning, the middle line of the ship points 
inside the circular arc and the thrust of the propellers is 
accordingly in a direction oblique to the path of the ship. 
The resistance to motion is therefore much greater than when 



3^4 



Theoretical Naval Architecture. 



on the straight, and this, together with the fore and aft com- 
ponent due to the rudder, EF in Fig. 130, causes a very 
considerable reduction of speed. In one case the reduction 
amounted to quite 50 per cent. 

If in Fig. 131 the ship is turning in the path GiGG 2 passing 
through the C.G., AF being the centre line of the ship and O the 
centre of the arc G 1 GG 2 , then the angle between the tangent 




GT and the centre line is termed the drift angle at the point G 
If OP is drawn perpendicular to the centre line there is no 
drift angle at P, and to an observer on board at P all points of 
the ship abaft P will appear to be moving to port, and all 
points forward of P will appear to be moving to starboard. 
Such a point P is termed a pivoting point, as the ship appears to 
pivot about P. 

The features of a ship which influence the turning are 
principally as follows : 

(a) Time taken to put the helm over to the maximum angle. 

() Pressure on the rudder. 

(f) Moment of resistance of the underwater body of the 
ship to turning. 



The Turning of Ships. 385 

(d) Moment of inertia of the vessel about a vertical axis 
through the centre of gravity, to which has to be 
added the mass of water associated with the ship. 

(a) The introduction of steam steering gear has rendered 
this item of less importance than formerly. In ships steered 
by manual power the time taken to put the helm over is 
considerable, and consequently the possibility of quick 
manoeuvring is small. 

The general adoption of balanced rudders has facilitated 
getting the helm over quickly, as the centre of pressure of the 
rudder is close to the axis and the moment required to be 
overcome is comparatively small. 

(b) The pressure on the rudder depends on various factors, 
which have already been dealt with above. 

(c) The ship when turning has angular velocity round the 
pivoting point P. If we take any portion of the ship at a 
distance /from P and of area A, the velocity through the water 

j/3 

due to the angular motion is I --71 and the resistance varies 

/ .jf\\ o 

as A f* - lj , and the moment of this about P varies as 

A /* ( -ft J . This therefore varies as the cube of the length 

and the square of the angular velocity. This moment at the 
early stages is small and less than the couple caused by the 
pressure on the rudder, and consequently the angular velocity 
increases. A point, however, is reached when the couple 
due to the pressure on the rudder is equal to the moment 
of resistance, and then the ship has a constant angular 
motion. 

It is seen by the above that if areas of the ship under 
water can be omitted where / is greatest, the resistance to 
the angular movement may be considerably reduced. This 
is done by the omission of the deadwood^ or the flat vertical 
portion of the ship aft, as the pivoting point is usually well 
forward. 

The above may be illustrated by the three turning circles 

2 C 



386 



Theoretical Naval Architecture. 



given in Fig. 132 of Orlando, Astrcea, and Arrogant. The 
profiles of the ships are given, from which it will be seen 
that 

(i) Orlando has a square type of rudder, not balanced, and 
the ship is 300 feet long. The former factor will 
delay her entry into the circular path. 

(ii) Astrcea has a balanced tudder, and the ship is 320 
feet long. The ship gets into the circular path 
quicker owing to the balanced rudder, but has a 
larger turning circle owing to the greater length. 

(iii) Arrogant. Here the rudder area is relatively large, 
two rudders being fitted. The length is 320 feet, 
as Astray but the stern is cut up considerably. 
The influence of these factors is seen in the very 
small circle, as compared with the Orlando of smaller 
length and the Astraa of the same length. 

Many merchant vessels now follow the practice of having 
balanced rudders with the deadwood aft cut away. In the 
Dreadnought the provision of two rudders with propellers 
immediately in front and the cut-up shape of the stern (as 
Fig. 144) resulted in a marked reduction of the turning circle. 
The following is the comparison with two cruisers of nearly 
the same length, one having a balanced rudder with no cut-up 
and the other having a balanced rudder with cut-up : 





















Tactical diameter. 




in feet. 


Rudder. 


Cut-up. 




In terms 










Yds. 


of ship's 












length. 


Powerful 


500 


Balanced 


("None, as\ 
I Fig. I37J 


IIOO 


6-6 


Duke of Edinburgh 


480 


Do. 


(As Fig.\ 
I id. 1 ? I 


740 


4*63 


Dreadnought ... 


490 


/ 2 No 

(Balanced 


V L ^J ) 

As Fig. j 
144 / 


463 


2-84 



The influence on the turning of a ship of a propeller acting 
directly on the rudder is strikingly illustrated by the comparative 



The Turning of Ships. 





FIG. 



132. 



383 



Theoretical Naval Architecture. 



turning circles of Topaze and Amethyst. These were cruisers 
of similar dimensions, viz. 360' X 40' x 14^' draught, 3000 

tons displacement. The 
shapes of the sterns 
are given in Fig. 133, 
from 
seen 
ship 



which it will be 
that the former 
was a twin-screw 
ship and the latter ship 
a triple - screw ship. 
This latter was the 
ship fitted with Par- 
sons' turbines, with 
small screws running 
screw immediately in 
of the two ships were 




FIG. 133. 
at high revolutions, and had one 



front of the rudder. The rudders 

of the same type, viz. balanced, and of much the same 

area. 

The turning circles are given in Fig. 134, from which it is 
seen that the twin-screw ship has a tactical diameter of 870 
yards and the triple-screw ship has a tactical diameter of 
550 yards, or 7*25 and 4-6 times the length of ship respec- 
tively. It is also seen that the latter ship gets into the circular 
path much sooner than the former ship. All the condi- 
tions are practically identical, except that the ship with the 
smaller circle has one propeller operating immediately on the 
rudder. 

(d) The moment of inertia of a ship about a vertical axis 
through the C.G. depends on the longitudinal distribution of 
the weight, which of course is decided upon for other reasons 
than turning. A ship with great weights at the ends will have 
a large moment of inertia, and a given turning moment due to 
the pressure on the rudder will take longer to get the ship into 
the circle than if the weights were more amidships. (It will 
be remembered that moment of inertia about any given axis 
is found by adding together the products of each portion of 
the wnight and the square of its distance from the axis, or 



The Turning of Ships. 



389 



Shapes of Sterns and Rudders. Fig. 135 shows the 
ordinary type of rudder fitted to merchant vessels. 

In some Atlantic and other liners what is termed the 
" cruiser type " of stern and rudder is adopted, analogous 




FIG. 134. 

to Fig. 143, the principal advantage being the increase 
of the length of the waterline obtained for a given length 
over all. 

Fig. 136 gives the stern and rudder adopted in the 
Aquitania. The normal shape of overhanging stern above 



390 



Theoretical Naval Architecture. 



water is obtained, but the deadwood is cut away and a 
balanced rudder obtained with the rudder-head below water. 
This gives the important advantage of having the rudder-head 
and steering gear under water and less liable to damage due 
to gun-fire. The vessel was built to be an auxiliary cruiser in 
case of necessity. 




FIG. 135. 



FIG. 136. 



Figs. 137 to 145 give a number of different shapes of 
sterns and rudders adopted in war vessels. 

Fig- J 37 was adopted for many vessels, including the 
protected cruisers Powerful and Terrible. The weight of the 
rudder is taken inside the ship, a steadying pintle only being 
provided at the bottom of the sternpost. 

Fig. 138 was adopted in the Arrogant class, designed as 
cruisers to company with the Fleet and in which exceptional 
turning facilities were desired. Two rudders are employed, and 
the deadwood is cut away. 

Fig. 139 is on similar lines with a single rudder. 

Fig. 140 is for the Japanese battleship Yas/iima, which had 
very great facility for turning. 

Fig. 141 was the type of stern adopted for many battle- 
ships ; the rudder is approximately square, and is unbalanced. 
The deadwood is cut away and the sternpost brought down to 
take the blocks for docking. 

Fig. 142 is the stern adopted in the King Edward VII, 
class, the rudder being partially balanced. 

Fig. 143 is the stern of the Lord Nelson class, similar to 
that of the Yashima. 



392 



Theoretical Naval Architecture. 



Figs. 144 and 145 show the double rudders employed in 
most of the Dreadnought battleships and battle cruisers. The 
rudder area by this means was made relatively large, and good 
powers of turning resulted, in spite of the greatly increased 




8A.TTLE -CRUISER 
FIGS. 143, 144, 145. 

length as compared with previous ships. In recent ships two 
rudders on the lines of Fig. 138 have been adopted because 
of the resistance caused by the bossing out to take the twin 
rudders. 

Turning Trials. It is usual to carry out systematic 
turning trials on H.M. ships, and these are put on record for 
the information of the officers. 



The Turning of Ships. 



393 



The following is the method employed to determine the 
path of the ship when turning. Two points are selected near 
the ends of the ship, at a known distance apart, and at these 
positions a horizontal circle graduated in degrees, etc., is set 
up with a pointer moveable in a horizontal plane, having sights 
which can be kept bearing on any given object as the vessel 
swings round. Two weighted rafts with a flag attachment are 
dropped overboard about a mile or so apart; it is assumed 
that these rafts remain stationary relative to the ship. 

The ship is brought up to one of these rafts, at the speed 
desired, so as to pass the raft as nearly as can be judged at 

L- 

I 

1 



TACT i CAL 



Di AM E 




FIG. 146. 



the distance of the radius of the turning circle expected away. 
Shortly before coming broadside on, a signal is made, when 
the rudder is put over, the course is noted and the time is 



394 Theoretical Naval Architecture. 

taken, and also the angles shown at the positions forward and 
aft, viz. OAB, OB A (Fig. 146), are recorded. These angles 
with the known distance AB fully determine the triangle OAB 
and consequently the position of the ship relative to the raft. 
A similar signal is made at four points (45), eight points (90), 
etc., and corresponding observations taken until the ship has 
completed the circle. The nine triangles found from this 
information are then set out on a convenient scale, as shown 
in Fig. 146, and the path of the ship drawn in. The " tactical 
diameter " and the " advance " can then be measured off. 

Angle of Heel when turning. On first putting a 
rudder over, the force on the rudder being usually below the 
centre of pressure on the hull on the opposite side, the resultant 
couple will have a tendency to heel the ship inwards, but this 
tendency is of short duration, as when the ship gets into her 
circular path centrifugal action comes into play and an outward 
heel results. It is shown in Chap. V. that this heel is 
given by 

V 2 d 

sin 6 = 0-088 . -5 -qrp 
R GM 

where V.is speed in knots; 

R is radius of turning circle ; 

GM is metacentric height ; 

d is distance of centre of lateral resistance below the 

C.G. 

A ship, therefore, of high speed, small turning circle, and 
small metacentric height will be liable to heel considerably 
when turning at full speed. 

Strength of Rudder-heads. The formula used by the 
British Corporation is as follows : 

^=0-26 4/R . A . S 2 

where A is area up to water-line in square feet ; 

R is distance of the C.G. of the area from the pintles ; 
S is not to be less than IT knots in vessels of and over 

250 feet in length. 

In vessels of 100 feet, speed taken as 8 knots. Inter- 
mediate lengths at intermediate speeds in proportion. 



The Turning of Ships. 395 

Lloyd's Rules do not now give a formula, but give 
diameters of rudder-heads for speeds varying between 10 and 
2 2 knots for different values of A X R as denned above. 

Direct Method of determining the Diameter of 
Rudder-head. The normal pressure on a rudder of area A 
square feet at angle of helm 6 is usually assumed to be 

P in Ibs. = ri2 A . ir . sin 6 
= 3-2 A. V 2 .sin0 

where v and V is speed of water past the rudder in ft. per sec. 
and knots respectively. 

It is usual to allow a percentage on to the speed of the 
ship to allow for the slip of the screw, although at the stern of 
the ship there is the "frictional wake." About 10 per cent, 
probably is well on the safe side. V is therefore taken at 
ri times speed of ship. 

In addition to knowing the pressure, it is necessary to know 
the point at which the centre of pressure acts in order to find 
the twisting moment about the axis. At 35 the centre of 
pressure is taken at three-eighths the breadth from the leading 
edge for a rectangular rudder. For other shapes of rudder the 
area may be divided approximately into rectangles, or we may 
adopt the method given later by dividing into a number of 
strips. 

Having obtained the twisting moment (preferably in inch- 
tons), we equate to the formula 

T =&../.* 

where d is diameter of rudder in inches ; 

/ is factor of strength allowed, say 

4 tons for wrought iron, 

5 ,, cast steel, 

3 phosphor bronze. 
The following example will illustrate the method : 

A rudder is 243 square feet in area, and the centre of pressure is esti- 
mated to be 6' 1 2 feet abaft the centre of rudder-head at 35. If the speed 
of ship is 19 knots, estimate the diameter of the rudder-head if of cast 
Steel. 



396 Theoretical Naval Architecture, 

Pressure in tons = - x 243 X 2O'9 2 x 0*574 = 87 tons 

Twisting moment = 87x6-12x12 = 6389 inch- tons 

.'. T'S * 5 <** = 6389, taking/= 5 
from which d = 187 inches. 

Note. If such a rudder is assumed to be a square and supported by two 
pintles at the forward edge, one at the bottom and one half-way up, it 
can be shown that, where W is the total load and / the total depth, that 

Bending moment at head = 3 ! s . W . / 1 , .. f , 

. , , , ITT 7 >both of these are small 

mid-depth = T fo . W./j 

Force at head = y W 

,, centre pintle = }W 
lower pintle = ^ W 

The above is an example of where pure twisting only need 
be considered (as for Fig. 141), but there are other cases to 
consider 

" (i) Rudder-head fixed in direction at sternpost, and the 
lower part supported at the bottom (as in Figs. 137 
and 142). 

(ii) Rudder-head fixed in direction at sternpost, and rudder 
supported about half-way up, the bottom being free 
(as in Figs. 140 and 143). 

(iii) Rudder fixed in sternpost and the lower part un- 
supported (as in Figs. 144 and 145). 

In (i) and (ii) both bending and twisting come into play. 
In (iii) bending is the determining factor in calculating the 
diameter of the rudder-head. It is generally assumed that the 
sternpost holds the rudder-head fixed in position. This gives 
results well on the safe side. 

(1) For the case (i) above, if the rudder is regarded as 
a beam uniformly loaded, it may be shown that 

' Bending moment at upper end = J x load X depth 
Support at head = f X load 
heel = | X load. 

(2) For the case (ii) above, if the rudder is regarded as a 
beam loaded at lower half to twice the intensity of the upper 



The Turning of Ships. 397 

half (i.e. the rudder is assumed to be a square with the upper 
corner cut out), it may be shown that 

Bending moment at upper end = ^ X load x depth 

Support at head = ~ x load\ in opposite 
heel = H x load/ directions. 

(3) For the case (iii) above, the bending moment at the 
head is found by multiplying the pressure by the distance of the 
C.G. below the top. 

When both bending and twisting have to be considered, 
we equate -^ . TT . f . a* to the equivalent twisting moment, viz. 
M + V M 2 + T a , where M is the bending moment and T the 
twisting moment. 

It may happen that the astern conditions will be the 
determining factor, because then the centre of pressure is 
nearer the after edge of rudder and farther from the axis than 
when the ship is going ahead. The speed astern is usually 
taken at half the speed ahead. In any case, in designing a 
rudder, sections must be made at various places besides the 

p M 
rudder-head, and the formula - = -j- applied to determine the 

value of the stress p. 

In fixing the shape of a rudder it has to be borne in mind 
that at all angles the centre of pressure should be abaft the axis. 
For angles below 35 the value three-eighths from the leading 
edge does not apply. The following formula may be used, 
based on Joessel's experiments for rectangular plates of 
breadth b. 

C.P. from leading edge = 0^195 b -f 0*305 b sin 

The formula given above, viz. P = 1*12 . A . iP . sin 0, is 
known to be incorrect for small plates moving through water, 
and the matter was exhaustively considered by Mr. A. W. 
Johns, R.C.N.C., at the I.N.A. for 1904. The formula, how- 
ever, has been extensively employed for many years for rudder 
calculations with satisfactory results. It is to be observed 
that 

(i) A rudder does not get the full angle at once, so that 



393 



Theoretical Naval Architecture. 



the pressure is not in the nature of a shock. We, 
however, use a coefficient of strength giving a large 
factor of safety, as if this were the case. 
(2) By the time the rudder is over the speed of the ship 

suffers an appreciable check. 

Centre of Pressure, Calculation of. For rudder 
shapes other than rectangular it may be assumed that if the 
C.G. is x feet from the mid-breadth, the centre of pressure 
is x feet from the position it would have if the rudder were rect- 
angular. Preferably the following method may be employed : 
Horizontal ordinates are drawn as shown (Fig. 147), 
common interval h, and f the length from forward and after 




FIG. 147. 

edges is set out on each. Such points represent the centre of 
pressure of strips of the rudder at the ordinates. Curves of 
centre of pressure are drawn as shown. 

Simpson's Rules then are applied, as indicated in the 
following table : 



The Turning of Ships. 



399 











Ahead. 


Astern. 












Products 






Ver- 


Product 


No. 


Ordi- 
nate. 


S. M. 


for 
area. 


C.P. 

from 


Product 
for 


C.P. 
from 


Product 
for 


tical 
Lever. 


for 
moment. 










axis. 


moment. 


axis. 


moment. 






I 


/i 


I 


/i 


0, 


** 


*, 


^, 








2 


/2 


4 


4/2 


^2 


4^2/2 > 


^2> 


4^2 7g> 


I 


4/2 


3 


/3 


2 


2/3 


rt s 


etc. 


etc. 


etc. 


2 


4/3 


4 


/4 


4 


4/4 


4 








3 


1 2y 4 , 


5 


/5 


2 


2/5 


a s 








4 


etc. 


* 


J; 


4 

2 


' ?: 











i 




8 


/8 


4 


4/8 


a 








7 




9 


/9 


i 




^ 








8 





8, 

Area = S l X ^ X A 

c; 
C.P. astern = from axis 



S 2 S 3 S 4 

C.P. ahead = |- 2 from axis 
^i 

C.P. below top = |- 4 X h 



The size of the rudder-head at the steering gear cross-head 
will be determined by the maximum twisting moment ahead or 
astern. 

The above method may be employed to approximate to 
the position of the centre of pressure at smaller angles by the 
use of Joessel's formula given above, in order to ensure that 
at all angles the centre of pressure is abaft the axis. 



CHAPTER XL 
LA UNCHING CALCULA TIONS. 

BEFORE starting on these calculations it is necessary to 
estimate as closely as possible the launching weight of the 
ship, and also the position of the centre of gravity both 
vertically and longitudinally. The case of the Daphne, 
which capsized on the Clyde 1 on being launched, drew 
special attention to the necessity of providing sufficient 
stability in the launching condition. A ship in the launching 
condition has a light draught, great freeboard, and high 
position of the C.G. It is possible, by the use of the prin- 
ciples we have discussed at length, to approximate to the 
metacentric height, and if this is not considered sufficient, 
the ship should be ballasted to lower the centre of gravity. 
It has been suggested that a minimum G.M. of i foot should 
be provided in the launching condition. If the cross-curves 
of stability of the vessel have been made, it is possible very 
quickly to draw in the curve of stability in the launching 
condition, and in case of any doubt as to the stability, this 
should be done. 

It is necessary to prepare a set of launching curves in the 
case of large heavy ships, in order to see that there is (a) no 
tendency for the ship to " tip," i.e. to pivot about the after end 
of the ways (as in Fig. 148 *:), in which case damage would 
probably ensue ; and (b) to obtain a value for the force which 
comes on the fore poppets when the stern lifts. 

In Fig. 148 , if G is the position of the C.G. of the ship, 
and B the centre of buoyancy of the immersed portion, then 
assuming a height of tide that may safely be expected for 
launching 

the moment of the weight abaft the after end of ways = W x d 
buoyancy = w x d l 

1 See Engineering '(1883) for a report on the Daphne, by Sir E. J. Reed. 



Launching Calculations. 



401 



Then for different portions of the travel down the ways the 
values of d, w, and d' can be readily found and curves drawn 
as in Fig. 149, giving values of (W x d) and (w X d) on a 
base of distance travelled. The former will, of course, be a 
straight line, starting from the point where the C.G. is over 
the aft end of ways. This line should be below the other 
curve, and the minimum intercept between them is called the 

"Su/ 




FIG. 148. 

" margin against tipping." If it happens that the curves inter- 
sect, it shows that a tendency to tip exists, and either (a) the 
ways should be lengthened, or (b) ballast placed forward, both 
of which increase the travel required before G comes over the 
end of the ways. The buoyancy curve should be drawn out 
for various heights of tide, in order to know the minimum 
height of tide on which the ship could be safely launched. 
This is a point of importance in some shipyards where tides 
do not always rise as high as expected owing to adverse winds. 
Ships have been launched successfully which had a tipping 
moment, but owing to the speed of launching the danger space 
was safely passed ; but this is a risk that few would care to 
take. 

As the ship goes further down the ways a position is 
reached when the moment of the buoyancy about the fore 

2 D 



402 



Theoretical Naval Architecture. 



poppet equals the moment of the weight about the fore poppet. 
At this point the stern of the ship will begin to lift. If W and w 
be the values of the weight and buoyancy respectively at this 
point, then the weight W w, instead of being taken over 
the length of ways in contact, is concentrated at the fore 
poppets. This weight is localized over a short distance both 
on the ship and on the slip, and it is desirable to know its 
amount and the position on the slip where it will come. 

Values of w are obtained at various points of the travel, 
and two lines drawn on a base of travel giving values of 



2.000.0_Q 10.000 




i.ooo.qoo 5.0Q&-- 



300' 



W and w, the former being constant. Then if a and b be the 
distances of the C.G. and the C.B. from the fore poppet (as 
in Fig. 148 a) at any point of the travel 

moment of weight about fore poppet = W x a 
buoyancy = w X b 

Curves are then drawn as in Fig. 149, giving these moments 
on base of travel, and the point where these curves cross gives 
the position where the stern begins to lift, and the intercept 
between the curves of weight and buoyancy at this point, 
viz. aa, gives the weight on the fore poppets. In this case 



Launching Calculations. 403 

the weight on the fore poppets was 2500 tons, the launching 
weight being 9600 tons. 

The launching curves for H.M.S. Ocean are given in a paper 
by Mr. H. R. Champness, read before the Institution of 
Mechanical Engineers in 1899. In that case the weight of 
the ship was 7110 tons, and the weight on the fore poppets 
1320 tons. 

The internal shoring of the ship must be specially arranged 
for in the neighbourhood of the fore poppets, and the portion 
of the slip under them at the time the stern lifts must be made 
of sufficient strength to bear the concentrated weight. 

Variation of Pressure on the Ways. In addition to 
knowing the pressure per square foot when the vessel is on 
the slip, it is sometimes desirable to know how this pressure 
varies as the ship goes down the ways. It is quite possible 
that the pressure might be excessive, and the necessity of 
strengthening the slip or shoring the ship internally would 
have to be considered. 

The support of the ways at any point of the travel is 
W w, where W is the weight and w the buoyancy. From 
this the mean pressure P can be determined. The support of 
the ways must act at a distance x from the fore end, such that 

(W-w)x = W.a-w.d 

where a and b are the distances of the C.G. and C.B. respec- 
tively from the fore poppet. Let y be the distance of the 
centre of pressure from the way ends. There are three cases 
to consider (see Fig. 150). 

(1) If x lies between \l and /, where / is the length of 
surface of ways in contact. Then, knowing P the mean 
ordinate, and assuming that the curve of pressure is a straight 
line, P A and P F , the pressures at the after and forward ends, 
can be determined. (See Example 28, Chapter II.) \ix~\l 
or /, then the maximum pressure is 2P, and occurs at the 
forward or after end as the case may be. 

(2) If x is less than ^/, the distribution of pressure is 
assumed a straight line on a base = 3^, and maximum pressure 

. 2 W - w 
at fore poppet is f . . 



404 



Theoretical Naval Architecture. 



(3) If y is less than \l we have similarly the maximum 

W - w 
pressure at the after end of way = f . . 

By this means a curve of pressure at fore poppet may be 
obtained for all positions of the ship on the slip until the stern 




FIG. 150. 

lifts, P F , and similarly for the after ends of ways, P A , as in 

(4) Fig- IS - 

The maximum pressure at the after end of the ways thus 
calculated in one ship was 3^ tons per square foot as compared 
with the mean pressure before launching of 1*7 tons. 



Launching Calculations. 405 

The following papers may be consulted in regard to the 
launching of ships : 

H. R. Champness, Ocean " I.M.E.", 1899 
W. J. Luke, Lusitania " I.N.A.", 1907 

J. Smith, "I.N.A.", 1909 

A. Hiley, " I.N.A.", 1913 



Example. In a certain ship the length of sliding ways was 535 ft. 
and the breadth 5 ft. 4 ins. The launching weight was 9600 tons with 
C.G. estimated at 247-5 ft- forward of the after end of sliding ways. 
Calculate the mean pressure per square foot on the ways, and assuming the 
pressure to vary uniformly as above, calculate the pressure per square foot 
at the forward and after ends of sliding ways before launching. 

Ans. 1*68 tons j 1*3 tons; 2*0 tons. 

(Note. These two latter values are the starting points of the curves in 
(4) Fig. 150, the latter rising to a maximum value- of 3| tons after a 
travel of 250 ft. The former curve rises to a high (indeterminate) value 
when the stern lifts.) 



APPENDIX A 



Proof of Simpson's First Rule. Let the equation of the curve 
referred to the axes O.r, Oy, as in Fig. 35, p. 53, be 

y = a Q + a\x + a. 2 x 2 

a , a^a 2 being constants ; then the area of a narrow strip length y 
and breadth &x is 

y x A.r 

and the area required between x o and x = 2^ is the sum of all 
such strips between these limits. Considering the strips as being a 
small breadth A.r, we still do not take account of the small triangular 
pieces as BDE (see Fig. 12), but on proceeding to the limit, i.e. 
making the strips indefinitely narrow, these triangular areas dis- 
appear, and the expression for the area becomes, using the formula 
of the calculus 



y . dx 
o 
or, putting in the value for y given by the equation to the curve 



J o 
which equals 

which has to be evaluated between the limits x = 2^ and x o. 
The expression then becomes 

Now, suppose the area = AJ/J + By 2 + Cy 3 



using the equation to the curve and putting x = o, x = h and 
x 2h respectively, 
Area = a (A + B + C) + ^(B + 2C) + a 2 . W(B + 4Q . (2) 



408 Appendix. 

By a well-known principle of Algebra we can equate the 
coefficients of a , a^ and a 2 in (i) and (2), so that 

A + B + C = 2h 

K.h 

B . # 

from which A = 

so that the area required is 



which is Simpson's First Rule. 

It may be shown in a similar manner that Simpson's First 
Rule will integrate also a curve which is of the third degree, viz. 

y = a + a^ . x + a.j? -F a s . X s . . . . (3) 

Simpson's First Rule is thus seen to integrate correctly curves 
both of the second and third degree. It is always used, unless the 
conditions are such that its use is not possible. 

Proof of Simpson's Second Rule. This may be proved 
similarly to the above, assuming that the curve has the equation (3) 
above. 

Proof of the Five-eight Minus-one Rule. The area between 
y l and y 2 is given by 



I, 



Assuming this to be equal to A^ + B/ 2 + Cy 3 , substituting for 
and J 3 , and equating coefficients of , a^ and 2 , we find 



so that the area required is 
5^. 
The area between the ordinates y 2 and y z is 

WGto+W'-'/i) 

and adding together, the whole area is 



which is Simpson's First Rule. 

Proof of the Three-ten Minus-one Moment Rule (given 
on p. 58). Assume the equation of the curve is 
y - a 4- ,* 



Appendix. 



409 



Then the moment about \ 
end ordinate/ 



n 

J o 



= yx.dx 



= h\\a Q + \ajt + tf 2 ^ 2 )on integrating 
Let the moment = (Ajj + By 2 + Cy 3 ). 

Substituting for y lt jr zt y 9 the values found from the equation to 
the curve, and equating coefficients of a 0) a 1} a. 2) we get 

A = ft#, B = !#, C = -sVfc 2 

so that moment = ^ih\^y l 4- loja ~J 3 ) 

Simple Area Rule for Six Ordinates (for which neither the 
first nor the second rules can be used). 

This is obtained by using the Five-eight Rule for the ends, and 
Simpson's Second Rule for the middle portion, thus 



I 



A, If, If, II, If, A 

ff[o'4, i, i, i, i, 0-4] 

Proof of TchebychefF's Rule with Four Ordinates. The 
following is the proof in the case where four ordinates are employed. 
In solving the equations for eight or ten ordinates imaginary roots 
are obtained, but the figures obtained for four and five ordinates 
can be combined together for the two halves of the length giving 
the figures in the table on page 18. 




FIG. 151. 

Let the curve BC (Fig. 151) be a portion of a parabola whose 
equation referred to the base and the axis OY is 

y = a + !.*+ a 2 .x* + a z .x*+ a.x* . . (i) 
where a , a lt a v a 3 , and 4 are constants. 

(For 4 ordinates the curve is taken of the 4th degree.) 
( .1 ., ,. th ) 



4 IQ Appendix. 

Let 2.1 be the length of the area, and select the origin at the 
middle of the length. 

/"*' 
Then the area required = / y doc (2) 

J -i 



r- 

. . (3) 



j j / 

Now let this area = C x the sum of the 4 ordinates (4) 



= C x 

C 

.... (5) 



< a .... 

substituting forj/u^-j, etc., their values as given by the equation 
to the curve (i), and taking the ordinates symmetrical about OY. 
Equating coefficients of a^ a a , a in the equations (3) and (5), 



we have C = - and 



From these equations we find 

jfi = o'i876/ 

** = 07947/ 

which gives the positions of the ordinates such that the area 

= L-(yi + y-i + y-i + J- 2 ), * the summation of the ordinates 

4 

is multiplied by the length and divided by the number of ordinates. 
Displacement Sheet by TchebychefTs Rule. This method 
may be extended to rinding the volume of displacement of a ship, 
and a table may be employed similar to that on Table I., 1 and 
described in Chapter II. There does not appear to be any advan- 
tage in applying Tchebycheffs rule in a vertical direction, as the 
number of water-lines are few in number compared with the 
number of ordinates usually employed fore and aft ; and also by 
having the waterplanes spaced equally, the displacement and 
vertical position of the C.B. for the other water-planes can be 
determined. In the specimen table, therefore, given on Table II., 1 
Tchebycheffs rule is employed for the fore-and-aft integration, 
and Simpson's first rule for the vertical integration. The figures 
shown in thick type are the lengths of the semi-ordinates of the 
various water-lines spaced from amidships as indicated at the top 
of the sheet. These lengths added up give a function of the area 

1 To be found at the end of the book. 






Appendix. 41 1 

of each water-plane, as 2417 for the L.W.P. These functions are 
put through Simpson's rule 1 in a vertical direction, and the 
addition of these products gives a function of the displacement, 
viz. 1 802-25. This function, multiplied for both rules, etc., as 
shown, gives the displacement in tons, viz. 16,067 tons. 

This result is obtained in another way, as in the ordinary 
displacement sheet, and an excellent check is thus obtained on 
the correctness of the calculation. The semi-ordinates of the 
various sections are put through Simpson's rule, and functions of 
the areas of the sections are thus obtained, as 255*05 for the 
section numbered II. These functions are then simply added up, 
and the same result is obtained as before for the function of the 
displacement, viz. 1802*25. It will have been noticed that the 
ordinates at equal distances from the mid-length are brought 
together ; the reason for this will appear as we proceed. 

The position of the centre of buoyancy of the main portion 
with reference to the L.W.P. is obtained in the ordinary way. To 
obtain the position of the centre of buoyancy of the main portion 
with reference to the mid-length, we proceed as follows. The 
functions of areas I. and I A. are subtracted, giving 2*9, and so 
on for all the corresponding sections. These differences are 
multiplied by the proportion of the half-length at which the 
several ordinates are placed, and the addition of the products 
gives a function of the moment of the displacement about the 
mid-length. In this case, the function is 36*4755. This, multi- 
plied by the half-length and divided by the function of the dis- 
placement, 1802*25, gives the distance of the centre of buoyancy of 
the main portion abaft midships, 6*07 feet. 

The lower appendage is treated in the ordinary way, as shown 
in the left-hand portion of the table. The reason of this is that 
the equidistant sections of the ship are usually drawn in on the 
body plan for fairing purposes, and the areas below the lowest 
water-line can be readily calculated. The sections at the stations 
necessary for Tchebycheff's rule would not be placed on the body 
for this calculation, but the ordinates at the various water-lines 
would be read straight off the half-breadth plan. 

The summary to obtain the total displacement and position 
of the centre of buoyancy is prepared in the ordinary way, and 
needs no explanation. The result of this summary is to give 
the displacement as 16,900 tons, having the centre of buoyancy 
11*2 feet below the L.W.L. and 6*52 feet abaft midships. 

1 In this case, instead of i, 4, 2, 4, 2, 4, i, the halves of these are 
used, viz. , 2, i, 2, i, 2 , the multiplication by 2 being done at the end. 



412 Appendix. 

Transverse BM. To determine the moment of inertia of the 
L.W.P. about the middle line, we place the ordinates of the L.W.P. 
as shown, cube them and add the cubes, the result being 211,999. 
This is multiplied as shown, giving 8,479,926 as the moment of 
inertia of the main portion of the L.W.P. about the middle line. 
Adding for the after appendage, we obtain 8,480,976 as the moment 
of inertia of the L.W.P. about the middle line in foot-units. 

The distance between the centre of buoyancy and the transverse 

metacentre is given by^, or 

8,480,976 

= 14-34 feet 
16,900 x 35 

The transverse metacentre is accordingly 14-34 11*2 = 3-14 feet 
above the L.W.L. 

Longitudinal BM. The position of the centre of gravity of the 
main portion of the L.W.P. is obtained by taking the differences 
of corresponding ordinates of the L.W.P. and multiplying these 
differences by 0-0838, etc., as shown. The addition of these pro- 
ducts, 14*6244, treated as shown, gives 18*15 f et as tne distance of 
the centre of gravity of the main portion of the L.W.P. abaft 
amidships. Adding in the effect of the after appendage, we find 
that the area of the L.W.P. is 29,144 square feet, and the centre of 
flotation is 19*53 feet abaft amidships. 

To determine the position of the longitudinal metacentre, we 
need to find the moment of inertia of the L.W.P. about a trans- 
verse axis through the centre of flotation. This has to be done in 
several steps. First we determine the moment of inertia of the 
main portion about amidships. This is done by taking the sum of 
corresponding ordinates and multiplying these by (o'o838) 2 , (0*3 127) 2 , 
etc., or 0-007, 0-098, etc. The addition of the products, 50*5809, is 
multiplied by 2 for both sides, by $ for TchebychefFs rule and 
by (300)2, being the square of the half-length, because we only 
multiplied by the square of the fraction of the half-length the 
various ordinates are from amidships, and not by the squares of 
the actual distances. The result gives 546,273,720 in foot-units 
for the moment of inertia of the main portion of the L.W.P. about 
the midship ordinate. We add to this the moment of inertia 
of the after appendage about the midship ordinate, obtaining 
539,399,902 in foot-units. This is the moment of inertia of the 
L.W.P. about the midship ordinate. To obtain the moment of 
inertia of the L.W.P. about a transverse axis through the centre 
of flotation, we subtract the product of the area of the L.W.P. and 



Appendix. 413 

the square of the distance of the centre of flotation abaft amidships. 
The result is the moment of inertia of the L.W.P. about a trans- 
verse axis through the centre of flotation we want, and this divided 
by the volume of displacement gives the value of the longitudinal 
BM, 927 feet. 

The moment to change trim one inch is obtained in the 
ordinary way, assuming that the centre of gravity of the ship is 
in the L.W.L. and that the draught marks are placed at the 
perpendiculars. 

To obtain Cross Curves of Stability by means of the 
Integrator and using TchebychefF's Rule. The rule we have 
been considering can be used with the integrator to determine the 
ordinary cross curves of stability in just the same way as with 
Simpson's rules. In Chapter V. the process of the calculation 
necessary with the integrator is explained. This calculation may 
be considerably shortened if Tchebycheff's rule is used instead of 
Simpson's rule. Not only can fewer sections be used, but the 
integrator itself performs the summation. In this case a body 
plan must be prepared, showing the shape of the sections at the 
distances from amidships required by the rule. Take, for example, 
a vessel 480 feet long, for which by the ordinary method twenty- 
one sections would be necessary. By using this rule nine sections 
will be quite sufficient, and by reference to the table on p. 18 the 
sections must be placed the following distances forward and aft 
of amidships, viz. 40*3, 126*9, I 44' 2 218*8 feet respectively, the 
midship section being one of the nine sections. 

The multiplier to convert the area readings of the integrator 
employed into tons displacement was for this case 1*097, and to 
convert the moment readings into foot-tons of moment was 13*164. 
All that is necessary, then, having set the body plan to the required 
angle as in Fig. 79, is to pass round all the nine sections in turn 
up to the water-line you are dealing with, and put down the initial 
and final readings. We have, for example 

Area readings. 

Initial 14*198 

Final 25,397 



11,199 difference 

Displacement in tons = 11,199 x 1*097 
= 12,285 tons 



Appendix, 



Moment readings. 
.. 5215 



Initial 
Final 



301 difference 
Moment = 301 x 13*164 = 3962 foot-tons 



It will at once be seen, on comparison with the example given 
on p. 199, that there is a very great saving of work by using this 
method. The following table gives the whole of the calculation 
necessary to determine a cross curve for the above vessel, values 
of GZ being obtained at four draughts, viz. at the L.W.L., one 
W.L. above and two W.L.'s below : 



Number of W.L. 


Area 
reading. 


Differ- 
ence. 


Displace- 
ment.* 


Moment 
reading. 


Differ- 
ence. 


GZ 


Initial 


1,505 


__ 





5,136 


_ 


_ 


A. W.L. 


14,198 


12,693 


13,924 


5,215 


79 


0*07 


L.W.L. 


25,397 


11,199 


12,285 


5,516 


301 


0-32 


2 W.L. 


35,126 


9,729 


10,673 


6,373 


57 


ro6 


3 W.L. 


43,301 


8,175 


8,968 


7,261 


888 


1-30 



Displacement Sheet (used in Messrs. John Brown $ CoSs 
Drawing Office}. The displacement table used in the drawing 
office of Messrs. John Brown & Co., Clydebank, presents several 
points of interest, and is admirably designed to conveniently con- 
tain on one sheet all the calculations necessary for the geometrical 
features of a ship's lines. This table was devised by Mr. John 
Black, and I am indebted to Mr. W. J. Luke for permission to 
reproduce it. The sections are numbered from aft, and the 
water-planes from below (see Tables III. and I HA. at end of 
book). 

. In Tchebycheff's three-ordinate rule (p. 18), the distance of the 
ordinates either side of the middle ordinate is 0707 times the half 
length of base. If, therefore, we apply this rule five times over for 
the length, we should set off on either side of Nos. i, 3, 5, 7, 9 
(Fig. 152) a distance of 0*0707 x length, the length being divided 
into ten equal parts. The addition of the ordinates A, B, C, . . . 



Appendix. 



4*5 



O, P, Q, multiplied by ^ length, gives the half area of the water- 
plane. 

The ordinates B, E, H, M, P (Fig. 152) are the following 

.distances from amidships (6, 3, o, 3, 6) . The ordinates A, C, D, 
F, G, K, L, N, O, Q, are the following distances from amidships, 
viz. (7-06, 4-94, 4-06, 1-94, ro6, ro6, 1-94, 4-09, 4-94, 7*06) . 

These are so close to the integers 7, 5, etc., that without appreci- 
able error the integers 7, 6, 5, 4, etc., can be used for the levers as 




in the ordinary displacement sheet. Thus, for any water-line the 
addition of ordinates in column A multiplied by 2 X will give 
the area. The algebraic sum of column B, divided by the addition 
of column A and multiplied by , gives the distance of the centre 

of gravity of water- plane from o to 10 from midships. Column D 
is got by multiplying the figures in column B again by the levers, 
and the addition of the column properly multiplied leads to the 
longitudinal moment of inertia of water-plane about amidships. 
This has to be corrected for the after appendage (if any), and then 
transferred to the centre of flotation, as explained in Chapter IV. 
From this the longitudinal B.M. is readily obtained for the several 
water-planes. 

In columns C are placed the cubes of the ordinates in columns 
A, and the addition of these columns leads to the transverse 
moment of inertia of the water-planes, from which values of the 
transverse B.M. is obtained for the several water-planes. 

The lower appendage is treated by " Thomson's rule," l the 
sections used being those on the ordinary body plan. The multi- 
pliers are obtained as follows : 



1 It is understood that ordinary sections and Simpson's multipliers are 
now used in this sheet for the appendage. 



416 



Appendix. 








i 


1 


2 


3 


4 


5 


6 


7 


8 


9 


9i 


10 


Area i to 9 
Areas o to i, 9 to 10 
Whole area 
Twice area 
Area 


I 


1 


fl 


2 

I 


cm 1 -4-m M M 


2 

I 




2 

I 


2 

I 


i 

2 

I 




2 

I 


i 


{ 


! 



The vertical C.B. of the appendage is obtained by Morrish's 
rule, viz. M - + - J, where d is the depth of appendage, v is its 

volume, and a is the area of No. I W.P. 

In the combination table the results are grouped together to find 
the displacement and C.B. up to Nos. 3, 5, 7, and 9 waterplanes. 

The displacement is found, in the first place, to the moulded 
surface of the ship, as is usual outside the Admiralty service ; the 
area of wetted surface is obtained by the formula S = 1 7 L . D + 
y 
=r, and using a mean thickness of plating, the displacement of the 

plating is readily obtained, and thus the "full" displacement is 
obtained. 

From the results a series of curves as in Fig. 153 is 
readily constructed, on base of draught, of displacement, tons per 
inch, centres of flotation, transverse metacentre, longitudinal 
metacentre, vertical C.B. fore and aft C.B., moment to change 
trim one inch, area of midship section and area of wetted surface, 
and also the various coefficients. To avoid confusion, the curves 
of vertical C.B. and metacentres are measured from the axis marked 
L.W.L ; those of C.B. and C.F. abaft amidships are measured from 
the right boundary of the figure. All others are measured from the 
left boundary. The scales used are appended to all the curves. 
These curves, when once carefully drawn for a ship, are of great 
value as records of the features of the ship's form. 

Loss of Stability due to Grounding. When a ship is being 
docked, the shores cannot finally be set up until the keel takes the 
blocks all fore and aft. Until this happens there is a portion of the 
weight taken by the after block (supposing the ship is trimming by 
the stern), and this becomes a maximum immediately before the 
ship grounds all fore and aft. Before the shores are set up there 
is a considerable upward pressure at the keel which might be 
sufficient, under certain circumstances, to cause instability, and 
cases are on record in which a ship has fallen over when being dry- 
docked owing to this cause. 




sj.Honvaa 



2 E 



4i8 



Appendix, 



In Fig. 154 let the first diagram represent the ship, floating 
freely, having a small inclination. In the second diagram a portion 
of the weight, w say, is taken by the blocks. This is equal to the 




FIG. 154. 

displacement between the lines W'L' and W"L". If M, be the 
metacentre corresponding to the water-line W"L", then 



Moment of stiffness = {(W 



- w . OG} sin 6 



To find w we can proceed as follows : 

1. Accurately. Obtain the displacement and longitudinal 
position of C.B. when floating freely. At the instant of taking the 
blocks all along, the moment of buoyancy about after block = 
moment of weight about after block. This equals moment of 
buoyancy about after block when floating freely. 

Hence we place a profile of the ship on the line of blocks, and 
draw a series of water-lines parallel to the keel. For each of these 
calculate the displacement and the longitudinal C.B. Draw out on 
a scale of draught a curve giving the moment of buoyancy about 
the after block. Where this crosses the constant line of the 
moment of weight about after block will give the draught at 
which the ship will ground, and so the displacement. This 
deducted from the original displacement gives the pressure on the 
blocks, and from the above the stability under these conditions can 
be determined. In a ship with small metacentric height and large 
trim by the stern, we have a combination of circumstances which 
would probably cause instability. The course to pursue is to keep 
the ship under control while any weight is taken by the blocks. 

2. Approximately. Suppose the vessel trims / feet by the stern, 



Appendix. 419 

and let the after block be b feet from the centre of flotation. When 
the vessel is floating* freely, imagine a force Q is applied at the 
after block just sufficient to bring the vessel to an even keel. 

Q = I2 ' , where M is moment to change trim I inch. 

The upward force Q will decrease displacement, and the 
mean draught is reduced by + T~^r ^ eet T being tons per 
inch. Owing, however, to the change of trim, the mean draught is 
increased by -j f eet > where the centre of flotation is c feet abaft 

amidships. If x is the draught at fore end when floating freely, 
then the mean draught when just grounding is 

. t .c M./ 



Theory of the Integrator. This instrument, shown in dia- 
gram in Fig. 79, gives by using suitable multipliers to the results 
obtained 

Ci) Area of a closed figure, 
(li) Moment of a figure about a given axis, 
(iii) Moment of inertia of figure about a given axis, 
by tracing out the boundary of the figure with the pointer of the 
instrument. 

*cr N 

Sx' 




FIG. 155. 

In Fig. 155 let M be the closed figure and AB the axis, P is a 
pointer at the end of an arm PC which is rigidly attached to a 
circle CL. The centre C of this circle is constrained to move 
along the line AB. Gearing with L is another circle N, centre D, 
CD always being perpendicular to AB. At the end of a radius 
DE of the circle N is a recording wheel capable of rotating about 



420 Appendix. 

DE, and this wheel can only record movements perpendicular 
toDE. 

Suppose the ratio of the circles L and N is as n : i. Then 
for an angular movement 6 of L the wheel radius DE will move 
through nd, and if when PC is on AB, DE is at an angle a, then 
when PC is at 0, DE is at an angle < = n6 + a. 

In going from P to the consecutive point P' on the curve, 
separated by a longitudinal interval 8.*-, we have to consider the 
influence of the recording wheel of two separate movements of 
PC, viz. 

(i) that due to the angular motion of PC ; and 

(ii) that due to the horizontal transfer 5^' of the centre C 
along AB. 

Consider now the influence of these two movements on the 
wheel 

(i) Since the curve is a closed curve the net result of the 

angular movement is zero. 

(ii) The recording wheel moves $x' parallel to AB, and the 
record on the wheel, i.e. its movement perpendicular to 
DE, is 5^. cos 0, 

= 5-r'. cos (n6 + a), 
and the total record 

= /cos (nB + a) . dx 1 

But 5.r = 5.r' + CP . 50 . sin 6. 

Hence the total record 

= /cos (nd + o) (8-r - CP . sin 6 . d&] 

= /cos (nB + a) dx. /CP . cos (n& + a) . sin 6 . d&. 

CASE I. Take n - i, a = 



The reading is /sin 6 . dx /CP . sin 2 . d& . 

The second term vanishes for a complete circuit, and since 
ordinate of the curve is CP sin 6, the reading is proportional to 
jy . dx, and therefore to the area. 

CASE 2 Take n = 2, a = o. 

The reading is /cos 26 . dx + vanishing terms, 

= /(i 2 sin 2 6) dx 

= fdx - 2 /sin 2 6 . dx 



- ^ jy 2 . dx, since \dx- 



Appendix. 421 

i.e. the reading is proportional to the moment of the area about 
AB. 

CASE 3. Take n 3, a -- . 
The reading is Jsin 3$ . dx 4- vanishing terms 
= J(3 sin 6 4 si 



the first term of which is proportional to area and the second 
term to the moment of inertia. 

Case 3 is little used in ship work. The student on first taking 
up the use of the integrator is advised to take simple geometrical 
figures of which the exact area and moment are known, and by 
this means the accuracy of the instrument may be tested, and, if 
necessary, any corrections made. 



MISCELLANEOUS EXAMPLES. 

1. The tons per inch immersion in salt water at a ship's water-planes 
are as follows, commencing with the L.W. P. : 12*9, 12*4, 11*5, 10*2, 8 - o, 
6*0, 2*2. The first five water-planes are 21 inches apart, and the last three 
are loj inches apart, the draught being 9 feet to bottom of flat keel. 

(a) Determine the displacement and the vertical position of the centre 
of buoyancy to the first three water-planes. 

(b) Estimate the displacement of the vessel when floating at a draught 
of 10 feet I J inches in water of which i cubic foot weighs 63! Ibs. 

Ans. (a) 1063 tons, 377 feet below L.W.L. 
797 473 
545 572 
(b) 1228 tons. 

2. Construct a formula giving the additional displacement, due to 
I foot greater trim aft as compared with the normal trim, in terms of the 
tons per inch immersion, length between draught-marks, and the distance 
of the centre of flotation abaft midships. 

The vessel in question, No. I, whose normal draught is 9 feet on an 
even keel, floats in salt water at a draught of 8 feet 7 inches forward and 
9 feet 10 inches aft. Estimate the displacement in tons, the centre of 
flotation being 7 feet abaft amidships, and the length P.P. 250 feet (draught- 
marks at perpendiculars). 

Ans. 12 =-^-, i loo tons. 
LJ 

3. The vessel in question No. I floats at a mean draught of 9 feet 
6^ inches in salt water. While in this condition she is inclined, two 
plumb-bobs 10 feet long being employed. The following deflections are 
observed : 



422 Appendix. 

Forward. Aft. 

3 tons through 23 ft. P to S 3 '6" ... 3-5" 

6 7* I 5" 7'05" 

Weights restored, ship came to upright. 

3 tons through 23 ft. S to P 3'55" ... 3'6" 

6 ,, ,, ,, 7' 1 5" " 7' 1 " 

Estimate the metacentric height at the time. 

Ans. 2 feet. 

4. A vessel of box form, 150 feet long and 25 feet broad, floats at an 
even draught of 8 feet, and has a water-tight deck 8 feet above keel. If 
a central compartment, 30 feet long, bounded by two transverse bulkheads 
extending up to the deck, is bilged, what will be (l) the new draught of 
the vessel ; (2) the alteration of the metacentric height ? 

Ans. 9' 8J" nearly ; increase nearly I foot. 

5. A body with vertical sides, the plan being an isosceles triangle 150 
feet long and 30 feet broad at tLc stern, floats in salt water at a constant 
draught of 10 feet. Determine the displacement when floating at a 
draught of 9 feet 6 inches forward, 10 feet 6 inches aft 

(a) by using formula obtained in question (2) above ; 
(6) by direct calculation, thus verifying (a). 

6. Obtain a rule for finding the area between two consecutive equi- 
distant ordinates of a curve when three are given. Show that the rule. 

when used with levers, results in a moment error of X (intercept between 

whole curve and chord), where h is the common interval. 

7. The half-ordinates of the water-plane of a ship 320 feet long and of 
9500 tons displacement are 1*0, 16-5, 25*0, 29*0, 30-4, 306, 30*5, 29*8, 
28*1, 24*1, and 15*1 feet. Find the sinkage of the vessel on passing from 
the Nore to the London Docks (63 Ibs. to cubic foot). 

Ans. 3*9 inches. 

8. If the vessel in the last question draws F 24' 3", A 27' 9" when at 
the Nore, find her draughts forward and aft when in the docks, the centre 
of buoyancy being 5*1 feet abaft middle ordinate and n feet below the 
centre of gravity. 

Ans. F 24' 7*", A 28' oj". 

9. A cigar-shaped vessel with circular sections floats in salt water with 
its axis in the surface. The semi-ordinates of the water-plane, 20 feet apart, 
are, commencing from forward, o, 3, 6, 8, 7, 4, I feet respectively. 

Find (i) Tons per inch immersion. 

(2) Displacement in tons. 

(3) Position of C.F from after end. 

(4) Position of C.B. ,, 

(5) Transverse BM. 

(6) Position of C.B. below W.L. 

Ans. (I) 276 tons; (2) 1577 tons; (3) 57 feet; (4) 567 feet; 

(5) 2-84 feet ; (6) 2-84 feet. 

Note. Some consideration should be given as to the simplest method 
of doing this question ; (6) should be inferred from (5). 

10. A vessel of constant triangular section is 245 feet long, 30 feet 
broad at the water-line, and floats at 12 feet draught with vertex downwards. 



Appendix. 423 

When a weight of 8 tons is moved 30 feet across the deck, a shift of 
8 inches is caused on the bob of a lO-feet pendulum. Find the position 
of the vessel's centre of gravity. 

Ans. 1 7 '64 feet from base. 

n. Assuming that a barge is of uniform rectangular section, 70 feet 
long and 20 feet broad, construct the metacentric diagram to scale for all 
draughts between 2 feet and 10 feet; state the draught for which the 
height of the metacentre above the keel is lowest, and show that in this 
condition the metacentre is in the corresponding water-plane. 

Ans. 8' 2". 

12. A right circular cone is formed of homogeneous material, and the 
tangent of the semi-vertical angle is O'5. Show that this cone will float in 
stable equilibrium with vertex down in fresh water so long as the specific 
gravity of the material is greater than 0*5 1 2. 

13. A long triangular prism of homogeneous material having the same 
section as the above floats in fresh water with vertex down. Show that 
it will float in stable equilibrium so long as the specific gravity of the 
material is greater than 0*64. 

14. A lighter has a constant section 16 feet at the base, 20 feet across 
the deck,iand 10 feet deep. She floats in river water 35*7 cubic feet to the 
ton at a constant draught of 8 feet, the length being 80 feet. The C.G. 
when laden to this draught is 6 feet above the base. 

Determine the angle of heel caused by taking 5 tons of the cargo out, 
this cargo being at 6 feet from the base and 6 feet from middle line. 

Ans. 2j to 2\ degrees. 

15. What relation exists between the transverse and longitudinal 
stability of a wholly submerged body ? 

Discuss the question of submarine navigation from the point of view of 
longitudinal stability. 

1 6. A lighter with vertical sides is 132 feet long and 30 feet broad for 
a length amidships of 80 feet. The ends are formed of four circular arcs 
of 30 feet radius. The draught is 10 feet, and the C.G. at this draught is 
7^ feet from the bottom. Determine the metacentric height. 

Ans. 4-35 feet. 

17. Prove the rule for the distance of the centre of gravity of a hemi- 
sphere of radius a from the bounding plane, viz. jj . a, by finding the BM 
of a sphere floating with its centre in the surface of the water. (See 
question 17, p. 141.) 

1 8. A ship of length 320 feet, breadth 50 feet, mean draught 19 feet, 
has a displacement of 4400 tons. The tons per inch at the L.W.L. is 
27, BM is 1 1 feet, and GM is 2-5 feet. 

It is proposed to design on similar lines a ship with the dimensions 
length 330 leet, breadth 51 feet, mean draught 19! feet. If G is the same 
distance above the keel in both ships, what value of GM would you 
expect in the new ship ? 

(Use approximate formula on pp. 66 and in.) Ans. 2f to 3 feet. 

19. A vessel of 700 tons displacement has a freeboard to the upper 
deck of 6 feet. The C.G. is i foot above water, and the metacentre locus 
is horizontal. A sea breaking over the bulwarks causes a rectangular area 



424 Appendix. 

50 feet long and 20 feet wide on the upper deck to be covered with water 
to a depth of I foot. Calculate the loss of metacentric height 

Ans. 1-5 foot. 

20. Show that for a vessel wall-sided in the neighbourhood of the 
water-line, GZ = (GM + JBM tan 2 0) sin at the angle of heel 9. 

Use this formula to determine the metacentric height in the upright 
condition of a box-shaped vessel, 200' X 35' X 10' draught, which is found 
to loll over to an angle of 5 (see p. 173). Ans. 0-04 foot. 

21. A tank, extending across an oil-carrying vessel, is 35 feet wide, 40 
feet long, and 10 feet deep. It has an expansion trunk at the middle line 
4 feet wide and 6 feet long. The vessel has a displacement of 2000 tons 
in salt water, and a GM of 2$ feet, the C.G. being 10 feet above the 
bottom of the tank. 

Find the virtual metacentric height when the tank is half full and also 
when filled. The density of the oil is 0*8 as compared with sea- water, 
and the metacentric curve is horizontal. 

Ans. (i) 1-54 foot ; (2) 3-19 feet. 

22. y lt y v y tt y^ y s , and y t are six consecutive equidistant ordinates 
of a plane curve : obtain the following expression for the area A of the 
curve lying between y l and y t , h being the common interval : 

A = 



23. A foreign vessel, whose form is not known, has a certain draught 
at the Nore, the sea-water there being 64 Ibs. per cubic foot. Off Green- 
wich, the water there being 63 Ibs. per cubic foot, it is noted that when 
loo tons have been unshipped the draught of water is again what it was 
at the Nore. What is the sea-going displacement of the vessel ? 

Ans. 6300 tons. 

24. A vessel 60 feet broad at water-line has the transverse metacentre 
12 feet above C.B., the latter being 10 feet below water. Find the height 
of metacentre above this water-line when 

(a) The beam is increased to 62 feet at the water-line, and in this 
ratio throughout, the draught being unaltered ; 

() The breadths at water-line are increased as above, but the lines 
fined so as to maintain the original displacement and to raise the 
C.B. 0-4 foot. Ans. (a) 2'8 feet ; (b) 3-64 feet. 

25. In a vessel whose moment to change trim one inch is M, tons per 
inch is T, and centre of flotation from after perpendicular is e times the 
length between perpendiculars, show that the position for an added weight 

such that the draught aft shall remain constant is - feet forward of the 

centre of flotation, and thus if the C.F. is at mid-length this distance is 

M 
2 . feet, or, approximately, one-ninth the length in a ship of ordinary 

form. 

26. Show that the distance forward of the after perpendicular at which 
a weight must be added so that the draught aft shall remain constant is 
given by moment of inertia of water-plane about the A. P. divided by the 
moment of water-plane about the A. P. 

27. A long body of specific gravity 0*5 of homogeneous material floats 
in fresh water, and has a constant section of the quadrant of a circle of 
10 feet radius. Determine the metacentric height when (a) corner upwards, 



Appendix. 425 

(b) corner downwards, and (c) when between the positions (a) and (b). 
Draw the general shape of the siability curve from zero to 360, starting 

with the body corner upwards. (The C.G. of the quadrant is times the 

radius from each of the bounding radii. Positions of stable and unstable 
equilibrium occur alternately.) 

Ans. (a} +2-33 feet ; (b) +2-33 feet ; (c) -0-36 foot. 

28. Two ships of unequal size are made from the same model. Prove 
that at the speed at which the resistance varies as the sixth power of the 
speed, the same effective horse-power is required for both ships at the same 
speed. 

29. A vessel 375 feet between perpendiculars is designed to float at 
21 feet F.P., 23 feet A. P. At this draught the displacement is 6500 tons 
salt water, tons per inch 45, and centre of flotation 13$ feet abaft amidships. 

The draught marks are placed on the ship 25 feet abaft the F.P. and 
35 feet before the A. P. respectively. Estimate as closely as you can the 
displacement when the draught marks are observed at the ship, 19' 6" 
forward, 23' 10" aft, when floating in water of which 357 cubic feet 
weigh I ton. 

Ans. 6293 tons. 

30. H.M.S. Pelorus is 300' X 36$' X 13$' mean draught, 2135 tons 
displacement, and requires 7000 I.H.P. for 20 knots. On this basis 
estimate the I.H.P. required for a vessel of similar form, 325' X 40' X 15!' 
mean draught, 3000 tons displacement, at 21 knots speed. State clearly 
the assumptions you make in your estimate. 

Among others the following assumptions are made : 

(1) For increased displacement caused by bodily sinkage, the I.H.P. varies as 

displacement for the same speed. 

(2) At the speeds mentioned in question, the I.H.P. is varying as the fourth power 

of the speed. 

Ans. About 10,600 I.H.P. 

31. The following formula has been proposed for the E.H.P. of a 
vessel at speed V knots, viz. 

E.H.P. = ^ |/. S . (V) 2 -" + b . ( ^- V'J 

where S = wetted surface in square feet. 
W = displacement in tons. 
L = length in feet. 
/ = a coefficient for surface friction. 
b = a coefficient varying with the type of ship. 

A vessel 500' X 70' X 26^', draught 14,000 tons, is tried at progressive 
speeds, and the curve of I.H.P. on base of speed shows the following 
values, viz. at 10, 12, 14, 16, 18, 20 knots, the I.H.P. is 1800, 3100, 
5000, 7500, 11,000, 15,500 respectively. 

Assuming the above formula to correctly give the E.H.P., determine 
the propulsive coefficients at the six speeds given. 

(Take/= 0-009, * = ' 2 S = 1 S'SJW x L as P- 262 ') 

Ans. (10) 46-4%; (12) 47'i / ; d4) 47'3/ ; (i6) 4 8-5 / ; 
(i8)48'9/o5 (20)49-8%. 

32. Using the above formula for E.H.P. (with /= 0-009, b = o'2s), 
determine the I.H.P. for speeds of 18 and 19! knots respectively of a 



Appendix. 

vessel 350' x 53^' x 20' x 5600 tons, using propulsive coefficients of 
45 / and 47* / respectively. 

Ans. 7660 I.H.P. ; 9740 I.H.P. 

33. Draw out the metacentric diagram for all draughts of a square log 
of 2 feet side, floating with one corner down. 

Supposing the log to be homogeneous, determine the limits between 
which the density must be in order that it shall float thus in stable equi- 
librium in fresh water. 

Ans. Between 0*28 and 0*72. 

34. A rectangular vessel is 175 feet long, 30 feet broad, 20 feet deep, 
and floats at a draught of 8 feet, with a metacentric height of 5 feet. Find 
the draught forward and aft, and the metacentric height due to flooding 
an empty compartment between bulkheads 120 feet and 150 feet from the 
after end. Ans. F. 13' 4!" ; A. 6' 7f" ; 4-2'. 

35. In a wall-sided vessel, show that for an angle of heel 6 the co- 
ordinates of the C.B. referred to axes through the C.B. in the upright 
condition are x = BM, .tan 9 ; y = $BM . tan 2 6. (BM, refers to the 
upright condition.) 

36. Using the above, show that a wall-sided vessel will heel to angle 6 
by shifting a weight w a distance d across the deck, 6 being given by the 
equation 



the suffix o referring to the upright condition. 

Thus, for a zero metacentric height the heel 9 is given by 

Y wxd 
tan = 2 . 



37. Show that a wall-sided ship having an initial negative metacentric 
height will heel to an angle of 9 = tan. / 2 GM Q, and will then have a 

metacentric height of 2GM . /i + 2 GM = 2 ^^. 
/ \/ BM ' cos 6 

38. Prove (by using BM = - J that the C.G. of a segment of a circle 
radius a, subtending an angle of 26 at the centre, is distant from the 

centre -a. : - : r - ^, and thus for a semicircle ( = - ) the C.G. 
3 v sin cos y 2j 

is - from centre. (Area of segment is area of sector less area of 

triangle, or a*0 a 2 sin 6 cos 6.) 

39. A vessel of 300 feet length floats at a draught of 12 feet forward, 
15 feet aft. (Tons per inch 18 ; moment to change trim I inch, 295 tons- 
feet ; C.F. 12 feet abaft midships.) It is desired to bring her to a draught 
not exceeding 12 feet forward and aft. How could this be done? 

Ans. Remove 350 tons 42^ feet abaft amidships (account is taken 
of increase of mean draught due to change of trim. ) 

40. Draw a curve of displacement for all draughts of a cylindrical 
vessel of diameter 20 feet and 150 feet long, and find, by using the curve, 
the distance of the C.B. from the base when floating at a draught of 
15 feet. 



Appendix. 427 

41. Draw the curve of displacement of a vessel of 14 feet draught 
having the following displacements up to water-lines 2 feet apart, viz. 
2118, 1682, 1270, 890, 553, 272, 71 tons, and by it find the position of 
the C.B. with reference to the top water-line. Suppose the tons per inch 
is 18*56, check your result by Morrish's formula. 

Ans. 5-45 feet. 

42. Prove the rule given on p. 19 for the volume of a sphere, by using 
Simpson's rules at ordinates, say, | the radius apart. 

(An exact result should be obtained, because the curve of areas is a 
parabola, which Simpson's rule correctly integrates.) 

43. A box-shaped vessel 140 feet long, 20 feet broad, 10 feet draught 
is inclined by shifting 7 tons 15 feet across the deck, and heels to an angle 
tan- 1 (J). Find the metacentric height (a) accurately, (b) by ordinary 
method. 

Ans. (a) 0-42 foot, (b) 0*52 foot. 

44. A long iron pontoon, of section 6 feet square and of uniform 
thickness, floats when empty in sea-water, but lolls over in fresh water. 
Find the thickness of the iron. (When the M curve is at mid-depth the 
draughts are 1-268 foot and 4732 feet. Of these the curve drops for 
increase of draught only in the former, so that for increase of draught as 
occurs in fresh water there is a negative metacentric height). 

Ans. J inch. 

45. A solid is formed of a right circular cylinder and a right circular 
cone of the same altitude h on opposite sides of a circular base radius r. 
It floats with the axis vertical, the whole of the cone and half the cylinder 

5 y4 21 

being immersed. Prove that the metacentric height is '~T~-jr-** 

so that for stable equilibrium r must be greater than 0*934^. 

46. A shallow-draught lightly built vessel is being launched. State 
the nature of the strains on the structure that will be experienced as she 
goes down. (From this point of view, the practice of some firms in 
launching torpedo-boat destroyers is of interest.) 

47. A vessel has a list to starboard due to negative metacentric height 
when upright. It is found that the addition of weights in the 'tween decks 
on the port side increases the list to starboard. How do you explain this ? 

48. If a swan or duck is floating in a pond, and reaches down to the 
bottom for food, why does the bird find it necessary to work with her feet 
to keep the head down and the tail up ? 

49. When floating in water, why is it necessary to keep the arms 
below? If the arms are raised out, what happens, and why? 

50. If a certain-sized tin is placed in water it will not float upright. 
When a certain quantity of water is poured in it floats upright in stable 
equilibrium. State fully the conditions of stability which lead to this 
result, bearing in mind the large loss of metacentric height due to the free 
surface of the water inside. 

51. A cube 12 inches side weighs 10-4 Ibs. Investigate the stability 
in fresh water, (a) with two faces horizontal, (b) with two faces only 
vertical and one edge downwards, (c) with a corner down. 

52. In going through the Caledonian Canal, the writer has noticed 
that the level of the water falls amidships. How do you account for this? 

53. If the water in question 36, Chap. III., goes right away with the 
tide, and the mud is very deep, investigate the stability of the vessel, 
the original metacentric height being 4 feet. 

Ans. Negative GM of foot. 



428 



Appendix. 



54. In a box-shaped vessel, 200 feet long, 30 feet wide, 10 feet draught, 
and having its C.G. li'Sfeet above the keel, a central transverse com- 
partment 50 feet long (assumed empty) is opened up to the sea. Will 
this vessel be stable after damage and free from danger in still water with 
a row of sidelights open, the lower edges of which are 1 5 feet above the 
keel? 

The GM when intact is 07 foot, and when damaged is O'5 foot, and 
new draught is 13,} feet, and in the final condition the vessel is all right, but 
there are intermediate conditions to consider, and the following table gives 
results of calculations with various depths of water in the middle com- 
partment : 



Height of water in 
compartment in !eet. 

I 
2 

3 
6 



Draught of 
water in feet. 



Metacentric height 
in feet 
-0-93 
-071 



-0-32 

croi 
+o'34 
+0-5 



Thus in the early stages the metacentric height is negative, and the 
ship will "loll," and if the hole through which the water enters is of 
comparatively small dimensions there would be an appreciable time for 
the list to develop. 

The above is taken from Prof. Welch's paper from the N.E. Coast 
Institution, 1915, on "The Time Element and Related Matters in some 
Ship Calculations," to which the reader is referred for a further develop- 
ment of the subject. 

SOLUTION OF QUESTION No. 21, CHAP. II., AND No. 36, CHAP. III. 

The author has had a number of requests as to the solution of these 
examples, and as they illustrate an important principle the solution is given 
below. 




FIG. 156. 

The metacentre when floating freely is readily obtained, viz. 16 feet 
from the base line. 

When the water level sinks 6 feet, the lower portion sinks into the 
mud, say x feet. Then, since the mud has a s.g. of 2, we take in the area 



Appendix. 429 

Owl twice. We equate the new displacement to the old, or 
(x + 6) 2 + * 2 = 144 

from which x = 4*94 feet. 

For a small inclination one half the buoyancy of the portion Ow/will 
act through m the metacentre of Owl, and the buoyancy of OWL' will 
act through M' the metacentre of OW'L', the portion Owl then being 
included twice. Om = 6'59' and OM' = 14*59'. The total buoyancy will 
act through a metacentre M such that 



(io'94 2 x I4'59) + (4'94 2 X 6-59) = I2 2 x OM, 
from which OM = 13 '2 feet. 

That is, the new metacentre is 2 '8 feet below the original metacentre, and 
as the C.G. of the ship has not been affected, the loss of metacentric height 
is 2| feet, about. 



APPENDIX B 

TABLES OF LOGARITHMS, SINES, COSINES, 
AND TANGENTS, SQUARES AND CUBES. 

LOGARITHMS. For some calculations considerable trouble is 
saved by using logarithms. One instance of this has been already 
given on p. 317. A table of logarithms is given on pp. 434, 435, 
to four places of decimals, which gives sufficient accuracy for 
ordinary purposes. To the right of the table are given the 
differences for I, 2, 3, etc., which enables the logarithms of 
numbers of four figures to be obtained. 

Thus log 2470 = 3*3927. The decimal part is obtained from 
the table, the whole number being 3, because 2470 is between 1000 
and 10,000 (log 1000 = 3, log 10,000 = 4). The log of 2473 is 
obtained by adding to the above log the difference in the table for 
3, viz. 5, i.e. 

log 2473 = 3*3932. 

The following are the principal relations in logarithms, viz. : 
log (M x N) = log M + log N 
log (jj) = log M -logN 
log (M) w = n . log M 
log VM = - . log M 

Thus multiplication is turned into addition, division is turned 
into subtraction, the raising to a power is turned into multiplication, 
and the taking of a root is turned into division. 

The decimal portion of a logarithm is always kept positive, and 



43 2 Appendix. 

the following are the values of the logarithm of the number 239 
for various positions of the decimal point : 
log 23,900 = 4-3784 
log 2,399 = 3-3784 
log 239 = 2-3784 
log 23-9 = 1-3784 
log 2-39 = 0-3784 
log 0-239 = - i + 0-3784 = [-3784 
log 0-0239 = - 2 + 0-3784 = 2-3784 
log 0-00239 = - 3 + o'3784 = 3'3784 
Example. To find the cube root of 10*75 : 
log 1075 = 1*0315 



log t/(iQ75) = J (1-0315) 

= 0-3438 

0-3438 = 0-3424 + 0-0014 
.*. V 10 '75 = 2*207 

Example. To find the value of (5725)$: 
log 5725 = 07578 

log (5725)* = f (07578) 
= 2-6523 

/. (5725)2 = 449'i 

Example.--!* find the value of <9 2 30 * > ('4'o8) 

5267 

log (9*31)1 = (3-9652) 

= 2-6435 
log (i4-o8) 3 = 3 (1-1485) 

= 3-4455 
log 5267 = 37216 

.-. Iog [(9*3.)3x( 14 -o8)*j = 2 6435 + 3-4455 - 37-6 



The number of which this is the log is 233 

. (9231)* x (I4'o8) 3 = 

5267 

Example. Find the value of 512 x 50*5 x 0*0037. 
log 512 = 27093 
log 50-5 = -7033 
log 0-0037 = 3-5682 
log (product) = 1-9808 
or product required = 95 66 



Appendix. 433 

Example. Find the value of Vo'oo765. 
log 0*00765 = 3*8837 

= 4 + 1-8837 

log Vo-00765 = i + 0*4709 
/. Vo'00765 = 0*2957 

NAPIERIAN OR HYPERBOLIC LOGARITHMS. 

These logarithms, which are also termed "natural," are 
calculated to the base e - 2718, and the following relation holds : 

log.N = 2*3 log 10 N 
Ordinary logarithms are calculated to the base 10. 

TABLE OF SINES, COSINES, AND TANGENTS. 

On pp. 436, 437, is given a table showing the values of the 
trigonometrical ratios, sines, cosines, and tangents, of angles up 
to 90, to three places of decimals, which will be found sufficiently 
accurate for ordinary purposes. 



2 F 



LOGARITHMS. 








1 


2 


3 


4 


5 


6 


7 


8 


9 


1 2 34 5 6\7 8 9 


10 

11 

12 
13 
14 


0000 

0414 
0792 
1139 

1461 


0043 

0453 
0828 

"73 
1492 


0086 
0492 
0864 
1206 
1523 


0128 

0531 
0899 
1239 
1553 


0170 
0569 

0934 
1271 
1584 


021202530294 
0607106450682 
0969100411038 

I303I335I367 
1614 1644 1673 


03340374 
07190755 
1072 1106 

1399 1430 
17031732 


4 8 12:17 21 25 
4 8 ii!i5 19 23 
3 7 1014 17 21 
3 6 10 13 16 19 
3 6 912 15 18 


29 33 37 
26 30 31 
24 28 31 
23 26 29 
21 24 27 


15 
16 
17 
18 
10 


1761 
2041 

2304 

2553 
2788 


1790 
2068 
2330 
2577 
2810 


1818 
2095 
2355 
2601 

2833 


1847 

2122 
2380 
2625 
2856 


1875 
2148 
2405 
2648 
2878 


1903 1931 

2175 22OI 

24302455 
2672 2695 
2900 2923 


1959 
2227 
2480 
2718 
2945 


19872014 
2253 2279 
25042529 
2742 2765 
2967 2989 


368 
3 5 8 

257 
2 5 7 
247 


n 14 17 
ii 13 16 

10 12 15 

9 12 14 
9 " 13 


20 22 25 

18 21 24 

17 20 22 

16 19 21 

16 18 20 


20 
21 
22 
23 

24 

25 
26 
27 
28 
20 

30 
31 
32 
33 
34 


3010 

3222 

3424 
3617 
3802 


3032 
3243 
3444 
3636 
3820 


3054 
3263 
3464 
3655 
3838 


3075 
328 4 

3483 
3674 
3856 


3096 
3304 
3502 
$692 

3874 


3Il8 

3324 
3522 

37" 
3892 


3139 
3345 
354i 
3729 
3909 


3160 

3365 
356o 
3747 
3927 


3181 3201 
3385 3404 
35793598 
37663784 
3945 3962 


246 
2 4 6 
2 4 6 
2 4 6 

245 


8 ii 13 

8 10 12 
8 10 12 

7 9 " 
7 9 ii 


15 17 19 

14 16 18 
M 15 17 
13 15 17 
12 14 16 


3979 
4150 
43M 
4472 
4624 


3997 
4166 
4330 
4487 

4639 


4014 
4i8^ 
4346 
4502 

4654 


4031 
4200 

4362 
4518 
4669 


4048 
4216 
4378 

4533 
4683 


4065 
4232 
4393 
4548 
4698 


4082 

4249 
4409 

4564 
4713 


4099 
4265 
4425 
4579 
4728 


4"64I33 
4281 429 

4440 4456 
45944609 

47424757 


2 3 5 
235 
2 3 5 
2 3 5 
i 3 4 


7 9 10 
7 8 10 
689 
689 
679 


12 I 4 15 
II 13 15 
II 13 14 
II 12 14 
10 12 13 


4771 
4914 

5051 
5185 

5315 


4786 
4928 
5065 
5198 
5328 


4800 
4942 
579 
5211 

5340 


4814 

4955 
5092 

5224 
5353 


4829 
4969 
5105 
5237 
5366 


4843 
4983 
5"9 
5250 
5378 


4857 
4997 
5132 
5263 
5391 


4871 
5011 
5H5 
5276 
5403 


48864900 
50245038 
5I595I72 
5289 5302 
54165428 


I 3 4 
i 3 4 
3 4 
3 4 
3 4 


679 
678 

5 7 8 
568 
568 


10 ii 13 

10 II 12 
9 II 12 
9 10 12 

9 10 ii 


35 
36 
37 
38 
30 


5441 
5563 
5682 

5798 
59" 


5453 
5575 
5 6 94 
5809 
5922 


5465 
5587 
5705 
5821 

5933 


5478 
5599 
5717 
5832 
5944 


5490 
5611 
5729 
5843 
5955 


5502 
5623 
5740 

5855 
5966 


5514 
5635 
5752 
5866 
5977 


5527 
5647 
5763 
5877 
5988 


5539 
5658 

5775 
5888 

5999 


5551 
5670 
5786 

5899 
6010 


I 2 4 
I 2 4 
I 2 3 
I 2 3 
I 2 3 


5 6 7 
5 6 7 

5 6 7 
5 6 7 
457 


9 10 ii 
8 10 ii 
8 9 10 
8 9 10 
8 9 10 


40 
41 
42 
43 
44 


6021 
6128 
6232 
6335 
6435 


6031 6042 
61386149 
6243 6253 

6345 6 355 
6444 6454 


6053 
6160 
6263 

6365 
6464 


6064 
6170 
6274 
6375 
6474 


60756085 
6i8o!6i9i 
62846294 

6385;6395 
64846493 


6096 
6201 
6304 
6405 
6503 


6107 
6212 
63H 
6415 
6513 


6117 
6222 

6325 
6425 
6522 


I 2 3 
I 2 3 

i 2 3 
i 2 3 

I 2 3 


4 5 6 
456 
4 5 6 
456 
4 5 6 


8 9 10 
7 8 9 
7 8 9 
789 
7 8 9 


45 
46 
47 
48 
40 


6532 
6628 
6721 
6812 
6902 


65426551 
66376646 

6730 6 739 
68216830 
69116920 


6561 
6656 
6749 

6839 
6928 


657i 
6665 
6758 
6848 
6937 


65806590 
6675 6684 
6767 6776 
6857 6866 
69466955 


6599 
6693 

6785 
f75 
6964 


6609 
6702 

6794 
6884 
6972 


6618 
6712 
6803 
6893 


i 2 31 4 5 6 
i 2 3 | 4 5 6 
i 2 3 | 4 5 5 
123445 
i 23445 


7 8 9 

7 7 ! 
678 
678 
678 


50 
51 
52 
53 
54 


6990 
7076 
7160 
7243 
7324 


6998 
7084 
7I68 1 
7251 
7332 


7007 
7093 

7U7 
7259 
7340 


7016 
7101 

7185 
7267 
7348 


7024 
7110 
7i93 
7275 
7356 


7033 7042 
7118 7126 
72027210 
7284 7292 
7364 7372 


7050 

7135 
7218 
7300 
7380 


7059 
7H3 
7226 
7308 
7388 


7067 
7152 
7235 
73i6 
7396 


i 23345 
123345 
122345 

i 22345 
122345 


678 
678 
6 7 7 
667 
667 



LOGARITHMS. 








1 


2 


3 


4 


5 


6 


7 


8 


9 


123 


456 


789 


55 
56 
57 
58 
59 


7404 
7482 
7559 
7634 
7709 


7412 
7490 
7566 
7642 
7716 


7419 
7497 
7574 
7649 
7723 


7427 

7505 
7582 

7657 
773i 


7435 
7513 
7589 
7664 
7738 


7443 
7520 
7597 
7672 

7745 


7451 
7528 
7604 
7679 
7752 


7459 
7536 
7612 
7686 
7760 


7466 

7543 
7619 
7694 
7767 


7474 
755 
7627 
770 
7774 


2 2 
2 2 
2 2 


3 4 5 
3 4 5 
3 4 5 
344 
344 


5 6 7 
5 6 7 
5 6 7 
5 6 7 
5 6 7 


60 
61 
62 
63 
64 


7782 
7853 
7924 
7993 
8062 


7789 
7860 

793 i 
8000 
8069 


7796 
7868 
7938 
8007 
8075 


7803 
7875 
7945 
8014 
8082 


7810 
7882 

7952 
8021 
8089 


7818 
7889 

7959 
8028 
8096 


7825 
7896 
7966 
8035 
8102 


7832 
7903 
7973 
8041 
8109 


7839 
7910 
798o 
8048 
8116 


7846 
7917 
7987 
8055 
8122 


2 


344 
344 
334 
334 
334 


566 
5 6 6 
5 6 6 
5 5 6 
5 5 6 


65 
66 
67 

68 
69 


8129 

f'95 
8261 
3325 
8388 


8136 
8202 
8267 
8331 
8395 


8142 
8209 
8274 
8338 
8401 


8149 
8215 

8280 

8344 
8407 


8156 
8222 
8287 

8351 
8414 


8162 

8228 
8293 
8357 
8420 


8169 
8235 
8299 

8363 
8426 


8176 
8241 
8306 
8370 
8432 


8182 
8248 
8312 
8376 
8439 


8189 
8254 

8319 
8382 

8445 




334 
334 
334 
334 
234 


5 5 6 
5 5 6 
5 5 6 
456 
456 


70 
71 
72 
73 
74 


8451 
8513 

8633 
8692 


8457 
8519 
*579 


8463 

8525 
8585 
8645 
8704 


8470 
8531 
8591 
8651 
8710 


8476 

8537 
8597 
8657 
8716 


8482 

8543 
8603 
8663 
8722 


8488 

8549 
8609 
8669 
8727 


8494 

fl 55 
8615 

8675 
8733 


8500 
8561 
8621 
8681 
8739 


8506 

8567 
8627 
8686 
8745 




234 
234 
234 
234 
234 


456 
455 

4 5 5 
455 
4 5 5 


75 
76 

77 
78 
79 


8751 
8808 
8865 
8921 
8976 


8756 
8814 
8871 

8927 
8982 


8762 
8820 
8876 
8932 
8987 


8768 
8825 
8882 
8938 
8993 


8774 
8831 
8887 

8943 
8998 


8779 
8837 
8893 

8949 
9004 


8785 

8842 
8899 
8954 
9009 


8791 
8848 
8904 
8960 
9015 


8797 

8854 
5910 

8965 
9020 


8802 
8859 
8915 
8971 
9025 


I 2 
2 
2 
2 
2 


233 
233 
233 
233 
233 


4 5 5 
4 5 5 

445 
445 
445 


30 

31 
32 

J4 


9031 
9085 
9138 
9191 
9243 


9036 
9090 

9H3 
9196 

9248 


9042 
9096 
9149 
9201 
9253 


9047 
9101 

9154 
9206 
9258 


9053 
9106 

9159 
9212 
9263 


9058 
9112 
9165 
9217 
9269 


9063 
9117 
9170 
9222 
9274 


9069 
9122 

9175 
9227 
9279 


9074 
9128 
9180 
9232 
9284 


9079 

9133 
9186 

9238 
9289 


2 
2 
2 

2 
2 


233 
233 
233 
233 
233 


445 
445 
445 
445 
445 


J5 
J6 

(8 
19 


9294 
9345 
9395 
9445 
9494 


9299 
9350 
9400 

945 
9499 


9304 
9355 
9405 
9455 
9504 


9309 
9360 
9410 
9460 
959 


9315 
9365 
9415 
9465 
9513 


9320 

9370 
9420 

9469 
9518 


9325 
9375 
9425 
9474 
9523 


9330 
938o 
9430 
9479 
9528 


9335 
9385 

9435 
9484 

9533 


9340 
9390 
9440 

9489 
9538 


2 
2 


233 
235 
223 
223 
223 


4 4 5 
4 4 5 
3 4'4 
344 
344 


'0 

'2 
3 
4 


9542 
9590 
9638 
9685 
9731 

9777 
9823 
9868 
9912 
9956 


9547 
9595 
9643 
9689 
9736 

9782 
9827 
9872 
9917 
9961 


9552 
9600 

9647 
9694 

974i 

9786 
9832 

9877 
9921 

9965 


9557 
9605 
9652 
9699 
9745 

9791 
9836 
9881 
9926 
9969 


9562 

9609 
9657 
J703 
9750 

9795 

9841 
9886 
9930 
9974 


9566 
9614 
9661 
97o8 
9754 

9800 

9845 
9890 

9934 
9978 


957i 
9619 
9666 

9713 
9759 

9805 
9850 
9894 
9939 
9983 


9576 
9624 
9671 
9717 
9763 

9809 

9854 
9899 

9943 
9987 


9628 

9675 
9722 
9768 

9814 
9859 

9903 
9948 
9991 


9586 

9633 
9680 

9727 
9773 




223 
223 
223 
223 
223 


3 4 A 
344 
344 
344 
344 

344 
344 
344 
344 
334 


5 
6 
7 
8 
9| 


9818 
9863 
9908 
9952 
9996 




223 
223 
223 
223 
223 





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TABLE OF SQUARES AND CUBES OF NUMBERS 
UP TO 50, RISING BY 0-05. 

The following table has been prepared, as squares and cubes 
of numbers are frequently required in ship calculations. Ordinates 
usually will not be measured more accurately than to the nearest 
0*05 ; in most cases the nearest decimal point is sufficiently accurate. 
The squares and cubes are taken to the nearest whole number, 
which is all that is necessary in ship calculations. 



, 


. 







4 










u 


| 




I 


u 




| 







s 


1 


Cubes. 


g 


a 

3 


Cubes. 


ifl 

S 


3 


Cubes. 


3 

fe 


0* 




fc 


JT 




3 
fe 


& 




0-05 


__ 


__ 


'55 


2 


4 


3-05 


9 


28 


O'lO 








60 


3 


4 


3*10 


10 


30 


0-15 








65 


3 


4 


3'!5 


10 


31 


0'20 








70 


3 


5 


3'2o 


IO 


33 


0-25 
0-30 





^_ 


32 


3 
3 


1 


3-25 
3-30 


II 
II 




o;35 








85 


3 


6 


3'35 


II 


38 










90 


4 


7 


3'4 


12 


39 


o'45 








95 


4 


7 


3|45 


12 


41 


0-50 








2'00 


4 


8 




12 


43 


0'55 








2-05 


4 


9 


3'55 


13 


45 


o'6o 








2'10 


4 


9 


3'bo 


13 


47 


0-65 








2-15 


5 


10 


3-65 


13 


49 


070 








2'20 


5 


ii 


370 


14 


51 


075 
0-80 




*~ ' 


2-25 
2-30 


5 
5 


ii 

12 


3-8^ 


H 


53 
55 


0-85 






2'35 


6 


13 


3-85 


15 


57 


0-90 






2-40 


6 


14 


3-90 


15 


59 


o'95 

I'OO 






2-45 
2-50 


6 
6 


II 


3'95 
4-00 


16 
16 


62 
64 


1-05 






2'55 


7 


17 


4 '05 


16 


66 


10 






2-60 


7 


18 


4-10 


17 


69 


15 




2 


2-65 


7 


19 


4-15 


17 




20 




2 


270 


7 


20 


4-20 


18 


74 


25 


2 


2 




8 


21 


4]25 


18 




30 


2 


2 


2'8o 


8 


22 




18 


80 


'35 


2 


2 


2-85 


8 


23 


4'35 


19 


82 


40 

45 


2 
2 


3 
3 


2*90 

2-95 


8 
9 


3 


4-40 
4'45 


19 

20 


8 5 

88 




2 


3 


3-00 


9 


27 


4-50 


20 


91 



Appendix. 



439 



I 


i 


Cubes. 


1 

I 


1 

1 

1 


Cubes. 


1 

fe 


t 

D 1 
U3 


Cubes. 


4'55 
4-60 
4-65 
470 

$1 


21 
21 

22 
22 
23 


94 

97 

101 

104 

107 


6'8o 
6-85 
6-90 

6'95 
7-00 


46 
47 
48 
48 
49 


314 
321 
329 
336 

343 


9'05 
9*10 

9'i5 
9-20 

9^5 


82 

83 

84 

11 




741 

754 
766 

779 
791 


4-85 
4-90 

4'95 
5-00 


*J 

24 
2 4 

25 
25 


114 
118 

121 
125 


7'05 
7-10 
7-15 
7-20 

7'2C 


50 
50 
5 1 
52 


350 

358 
366 
373 

7 o T 


o u 
9'35 
9-40 

9'45 
9'5o 


87 

88 

89 
90 


804 
817 
831 
844 
857 


5-05 
5-10 

5'i5 

5 '20 

5-25 


26 
26 
27 

3 

28 


I2 9 
133 
137 
I 4 I 

H5 


/ O 
7'30 

7'35 
7-40 

7'45 
7'5o 


53 
53 
54 

1 

56 


3 51 
389 
397 
405 
4 r 3 
422 


9-55 
9*60 

9-65 
970 

975 
o'8o 


9i 
92 
93 
94 
95 


871 
885 
899 
913 
927 


3 
5'35 
5-40 

5 '45 
5^ 


29 

29 

30 
30 


149 

'53 

'I 7 
162 

166 


7'55 
7 -60 
7-65 
770 

7*7C 


11 

59 

59 


430 

439 
448 

457 


9-85 
9-90 

9-95 

10 '00 


y 
97 
98 
99 

100 


941 
956 
970 

985 
1,000 


5'55 
5-60 

5'65 
57o 
575 

5.o n 


31 
31 

32 
32 

33 


171 

176 
180 
185 
190 


/J 
7'80 

7'85 
7-90 

7'95 
8-00 


61 
62 
62 

63 
64 


5 

475 
484 
493 
502 
512 


10-05 

lO'IO 
IOT5 
I0'20 

10-25 

JQ'OQ 


101 

1 02 
103 
104 

lol 


1,015 
1,030 
1,046 
1,061 
1,077 


50 

5^5 
5-90 

5'95 
6'oo 


34 
34 
35 
35 
36 


95 

200 

205 

211 

216 


8-05 
8-10 
8-15 
8-20 

8-2C 


65 
66 

66 
67 

AQ 


522 
531 
54i 
551 

Cfi2 


10*35 
10*40 
10-45 
10*50 


107 

108 
109 
no 


I >93 
1,109 

1,125 
1,141 

1,158 


6*05 
6'io 
6''5 

6*2O 
6-25 
6"2O 


37 
37 

38 
38 
39 


221 

227 

233 
23* 
244 


^ 

8-30 

8'35 
8-40 

8'45 
8-50 


69 
70 
7i 
7i 
72 


5O2 

572 
582 

593 
603 
614 


!0'55 
io'6o 
10*65 
10-70 

10-75 
10*80 


III 

112 

"3 

114 

116 


i,i74 
1,191 
1,208 
1,225 
1,242 


6-35 

6 40 
6-45 
6-50 


4U 
40 

41 
42 
4 2 


250 
256 
262 

268 

275 


8'55 
8-60 
8-65 
870 

8'7C 


73 

74 
75 
76 


625 
636 
647 
659 


10*85 
10-90 
10-95 

H'OO 


118 
119 
1 20 

121 


1,277 
i,295 
1,313 
i,33i 


6'55 
6-60 
6-65 
670 
675 


43 

44 
44 
45 
46 


281 
287 
294 
301 
308 


/D 
8'80 
8-85 
8-90 

8'95 
9 -oo 


77 
77 
78 
79 
80 
81 


681 
693 
705 
717 
729 


11-35 

II'IO 

11*15 

1 1 -20 

11-25 


122 

123 
124 

125 
I2 7 


1,349 
1,368 
1,386 
1,405 
1,424 



440 



Appendix. 



Numbers. 


t 

! 


Cubes. 


Numbers. 


1 


Cubes. 


Numbers 




1 
1 


Cubes. 


11-30 

1 1 '35 
1 1 '40 
11-45 
11-50 


128 
129 
130 
131 

132 


,443 
,462 
,482 
,501 
,521 


13-55 
13*60 

I3-65 
13-70 

1375 


184 
185 
1 86 
1 88 
189 


2,488 
2,515 
2,543 
2,57i 
2,600 
2fi-?8 


I5*8o 

I5-85 
15-90 

I5'95 

16*00 


250 
251 
253 
254 
256 


3,944 
3,982 
4,020 

4,058 
4,096 


"'55 
ir6o 
11*65 
11-70 


i33 
i35 

136 
137 

T 78 


,541 
,56i 
,581 
,602 
62? 


13 oO 

13-85 
13*90 

13-95 
14*00 


190 
192 
193 
195 
196 


2,657 
2,686 
2 ,7I5 
2,744 


16*05 
16*10 
16*15 
16*20 


258 
259 
261 
262 

*>f\A 


4,135 
4,173 

4,212 

4,252 


75 
11-80 
11-85 
11-90 
"'95 

12-00 


13 
139 
140 
142 
143 
144 


,O32 

,643 
,664 
,685 
,706 
,728 


14*05 
14*10 

14-15 
14*20 
14*25 


197 

199 

200 
202 
203 


2,774 
2,803 

2,833 
2,863 
2,894 


16*30 

16-35 
16*40 
16*45 
16-50 


266 
26 7 
26 9 
271 
272 


,291 
4,331 
4,371 
4,411 

4,451 
4,492 


I2-05 
12-10 
12-15 
12-20 
I2'2C 


145 
146 
148 
149 


,750 
,772 

,794 
,816 

C 7 


*4 o u 
14-35 
14*40 

14-45 
H-50 


206 
207 
209 
2IO 


xy*4 

2,955 
2,986 

3,oi7 
3,049 


16-55 
16*60 
16*65 
16*70 


274 
276 

277 
279 

C T 


4,533 
4,574 
4,616 

4,657 
A 600 


12 25 
I2-30 

I2'35 
12*40 

I2'45 
12*50 


I 5 
IS' 
153 
154 
155 
156 


>j 
,861 

,884 
,907 
,930 
,953 


H'55 
14*60 

!4"65 
14*70 

!4"75 
iA*8rj 


212 
213 

215 

216 
218 


3,o8o 

3,112 
3,144 
3,177 
3,209 


1U /i 
16*80 
16-85 
16*90 

16-95 
17*00 


282 
284 
286 
287 
289 


4,742 
4,784 
4,827 
4,870 
4,913 


I2'55 
12*60 
I2*65 
12*70 


158 

'59 
160 
161 

167 


i,977 
2,000 
2,024 
2,048 


14-85 
14*90 

14-95 
15*00 


51 

221 
222 
224 
225 


>^4 Z 

3,275 
3,308 

3,341 

3,375 


17-05 
I7'io 

I7-I5 
17-20 

I 7*2C 


291 
292 

294 
296 
298 


4,956 

5,000 

5,044 
5,o88 
51 33 


12 75 
12*80 
12*85 
12*90 
12*95 
13*00 


^ 3 
164 

165 
166 
1 68 
169 


2 ,73 
2,097 

2,122 
2,147 
2,172 
2,197 


I5-05 
15*10 

I5-I5 
15*20 

I5-25 


227 
228 
230 
231 
233 


3,409 

3,443 
3,477 
3,512 
3,547 

? rXo 


*/ *3 

17-30 

I7-35 
17*40 

I7'45 
I7*50 


299 
301 

303 
3j 
306 


, L O3 

5,178 
5,223 

5,268 
5,3H 
5,359 


13*05 
13-10 

13*15 

13-20 


170 
172 
173 

174 
11 f\ 


2,222 

. 2,248 
2,274 
2,300 


1 S 3 
I5-35 
15-40 
i5'45 
i5'5o 


234 
236 
237 
239 
240 


35 52 
3,6i7 
3,652 
3,688 
3,724 


17-55 
I7*6o 

17-65 
IT70 

I7"7 c 


308 
310 
312 
313 

71 C 


5,405 
5,452 
5,498 
5,545 

5CQ2 


*3 2 5 
I3"3o 

I3-35 
13-40 

13-45 
I3-50 


170 
177 
178 
1 80 
181 
182 


2,326 

2,353 
2,379 
2,406 

2,433 
2,460 


15-55 
15*60 

15-65 
15*70 

1575 


242 

243 
245 
246 
248 


3,76o 
3,796 
3,833 
3,870 
3,907 


/ IJ 

I7*80 

I7-85 
I7*90 

I7-95 

18*00 


6 1 J 

3 1 7 
3i9 
320 
322 
324 


oy- 6 
5,640 
5,687 
5,735 
5,784 
5,832 



Appendix. 



441 



1 

3 
fe 


I 


Cubes. 


1 

D 




! 

i 


Cubes. 


1 

& 




1 


Cubes. 


I8-05 

18-10 
18-15 
18-20 
18-25 




329 
331 

333 


5,881 

5.930 

5,979 
6,029 
6,078 
6 128 


20-30 

20-35 
20-40 
20-45 
26-50 


412 
414 
4 i6 
4 i8 
420 


8,365 
8,427 
8,490 
8,552 
8,615 


22-55 

22"6O 
22-65 
22-70 
22-75 

2?'8f> 


509 
5" 
5i3 

5i5 
5i8 
c?o 


11,467 

",543 
11,620 
11,697 

",775 
U8c? 


lo 30 

i8-35 
18-40 

18-45 
18-50 


335 
337 
339 
340 
342 


6,179 
6,230 
6,280 
6,332 


20-55 
20 '60 
20*65 
20-70 


422 

424 
426 
428 


8,678 
8,742 
8,806 
8,870 

8Q1A 


22-85 
22-90 

22-95 

23-00 


^zu 

522 
524 
527 
529 


n,930 
12,009 
12,088 
12,167 


i8'55 
18-60 
18-65 
18-70 
i8-75 

rC'SX 


344 
346 
348 
350 
352 


6,383 
6,435 
6,487 
6,539 
6,592 
ft 6/1 c 


20 75 
20-80 
20-85 
20-90 
20-95 

2I'OO 


43 1 
433 
435 
437 
439 
441 


,Vj4 

8,999 
9,064 
9,129 

9,195 
9,261 


23-05 

23-10 

23-15 

23-20 
23-25 


531 

534 
536 
538 
54i 


12,247 
12,326 
12,407 
12,487 
12,568 


15 oO 

I8-85 
iS^O 

18-95 
I9-00 


353 
355 
357 
359 
361 


, U 45 
6,698 

6,751 
6,805 

6,859 


21-05 
2I"IO 
21-15 
21-20 


443 
445 
447 
449 


9,327 
9,394 
9,461 
9,528 


23 30 

23-35 
23-40 

23-45 
23-50 


543 

545 
548 
550 
552 


12,649 

12,731 
12,813 
12,895 
12,978 


I9-05 
I9-IO 

I9-I5 
I9-20 
I9-25 


363 

3 ^ 5 
367 

369 

37i 


6,913 
6,968 
7,023 
7,078 

7,133 
7I&Q 


21 25 
2I-30 

21-35 

2I-40 

21-45 
21-50 


452 
454 
456 
458 
460 
462 


,59 
9,664 
9732 
9,800 
9,86 9 
9,938 


23-55 

23-60 

23-65 
23-70 

23-75 

o-7'Xn 


555 
557 
559 
562 

564 
-f.f. 


13,061 

13-144 
13,228 
13,312 
13,396 


I9-30 

I9-35 
I9-40 

19-45 
I9-50 


37 2 
374 
376 
378 
380 


,ioy 

7,245 
7,301 
7,358 
7,415 


2i'55 
21-60 
21*65 
21-70 

21 '7 


464 
467 
469 
47i 

A.T\ 


10,008 
10,078 
10,148 
10,218 


23-85 
23-90 

23-95 
24-00 


569 
57i 

574 
576 


13,40! 

13,566 
13,652 
13,738 
13,824 


I9-55 
I9-60 
I9-65 
I9-70 
1975 


382 
384 
386 
388 
390 


7,472 
7,530 
7,5^7 
7,645 
7,704 


21 75 

2 1 'SO 
2I-85 
2I-90 
21-95 
22'OO 


H-/O 

475 
477 
480 
482 
484 


lUj^oy 
10,360 
10,432 
10,503 
10,576 
10,648 


24-05 
24-10 

24-15 
24-20 

24-25 


578 
58i 
583 
586 
588 


I3,9H 
13,998 
14,085 
14,172 
14,261 


19 oO 

19-85 
19-90 
19-95 

20-00 


392 
394 
396 
398 
400 


7-j 
7,821 

7,88 1 
7,940 
8,000 


22-O5 
22'IQ 
22-15 
22-20 
22'?C 


486 
488 
491 
493 

/IOC 


10,721 

10,794 
10,867 
10,941 


24 3 
24-35 
24-40 

24-45 
24-50 


59 
593 
595 

& 


! 4>349 
14,438 
H,527 
14,616 
14,706 


2O'O5 
20-10 
20 15 
20-20 
2O-25 


402 
404 
406 
408 
410 


8,060 

8,121 

8,181 
8,242 
8,304 


22 25 
22-30 
22-35 
22*40 
22-45 
22*50 


47J 

497 
500 
502 
504 
.506 


11,015 
11,090 
11,164 
11,239 
",3i5 
n,39i 


24-55 
24-60 
24-65 
24-70 
24-75 


| 

608 
610 
613 


14,796 
14,887 
H,978 
15,069 
15,161 



442 



Appendix. 



Numbers. 


rt 

D 

cr 
M 


Cubes. 


Numbers. 


t 

m 
1 


Cubes. 


Numbers. 


t 



1 


Cubes. 


24-80 
24-85 
2490 

24-95 
25-00 


615 

618 
620 
623 
625 


15,253 
15,345 
15,438 
15,531 
15,625 


27-05 
27-10 

27-15 
27-20 
27-25 
27-3 
27-35 
27-40 
27-45 
27-50 


732 

734 
737 
740 
743 
745 
748 
75i 
754 
756 


19,793 
19,903 
20,013 
2O,I24 
20,235 
20,346 
20,458 
20,571 
20,684 

20,797 


29-30 

29-35 
29-40 

29H5 

29-50 


858 

861 
864 
867 
870 


25,154 
25,28 3 
25,412 
25'542 
25,672 


25-05 
25-10 

25-I5 
25-20 

25-25 
25-30 

25-35 
25-40 

25-45 
25-50 


628 
630 
633 
635 
638 
640 
6 43 
645 
648 
650 


15,719 
15,813 
15,908 
16,003 
16,098 
16,194 
16,290 

16,387 
16,484 
16,581 


29-55 

29-60 

2965 
2970 
2975 

29-80 

29-85 
29-90 
29-95 

30-00 


873 
876 

879 
882 
885 
888 
891 
894 
897 
900 


25,803 

25,934 
26,066 
26,198 

2 ^33I 
26,464 

26,597 
26,731 
26,865 
27,000 


27-55 
27-60 

27-65 
2770 

27-75 
27-80 
27-85 
27-90 

27-95 
28-00 


759 
762 
765 
767 
770 
773 
776 
778 
78i 
784 


20,911 
21,025 

21,139 
21,254 

21,369 
21,485 

2I,60T 
2I,7l8 
21,835 
21,952 


25'55 
25-60 
25-65 
25-70 

25-75 
25-80 

25-85 
25-90 

25-95 
26-00 


653 
655 
658 
660 
663 
666 
668 
671 

673 
676 


16,679 
16,777 
16,876 

16,975 
17,074 

17,174 
17,274 
17,374 
17,475 
17,576 


30-05 

30-10 

30-15 
30-20 
30-25 
30-30 
30-35 
30-40 

30-45 
30-50 


903 
906 

909 
912 

9i5 
918 
921 
924 
927 
930 


27,135 
27,271 

27,407 
27,544 
27,681 
27,8l8 
27,956 
28,094 
28,233 

28,373 


28-05 
28-10 
28-15 
28-20 
28-25 
28-30 

28-35 
28^40 
28-45 
28-50 


787 
790 
792 
795 
798 
80 1 
804 
807 
809 
812 


22,070 
22,188 
22,307 
22,426 

22,545 
22,665 
22,786 
22,906 
23,028 
23,H9 


26-05 
26-10 
26-15 
26-20 
26-25 
26-30 
26-35 
26-40 

26-45 
26-50 


679 

68 1 
684 
686 
689 
692 
694 
697 
700 
702 


17,678 
17,780 
17,882 

I7,985- 
18,088 
l8,I9I 
18,295 
18,400 
18,504 

18,610 


30-55 
30-60 

30-65 
30-70 

30-75 
30-80 

30-85 
30-90 

30-95 

31*00 


933 
936 
939 
942 
946 
949 
952 
955 

$ 


28,512 
28,653 

28,793 
28,934 
29,076 
29,218 
29,361 
29,504 
29,647 
29,791 


28-55 
28-60 
28-65 
28-70 

28-75 
28-80 
28-85 
28-90 
28-95 
29-00 


815 
818 
821 
824 
827 
829 
832 
835 
838 
841 


23,271 
23,394 
23,5!7 
23,640 
23,764 
23,888 
24,013 
24,138 
24,263 
24,389 


26-55 
26-60 
26-65 
26-70 

2675 
26-80 
26-85 
26-90 
26-95 
27-00 


705 
708 
710 

713 
716 
718 
721 
724 
726 
729 


18,715 
18,821 
18,927 

19,034 
19,141 
19,249 
19,357 
19,465 
19,574 
19,683 


31-05 
31-10 

31-15 
31-20 

31-25 
31-30 
31-35 
31-40 
31-45 
31-50 


964 
967 
970 
973 
977 
980 

983 
986 

989 
992 


29,935 
30,080 
30,226 

30,37i 
30,518 
30,664 
30,811 
30,959 
3M07 
31,256 


29-05 
29-10 
29-15 
29-20 
29-25 


844 
847 
850 

853 
856 


24,515 
24,642 

24,769 
24,897 
25,025 



Appendix. 



443 



J 


ui 
f 

I 


Cubes. 


jt 

a 

3 



i 
i 


Cubes. 


| 

1 


ui 

z 

rt 

3 

o* 
in 


Cubes. 


31-55 

31-60 

31-65 
31-70 

31-75 

2 1 '80 


995 
999 

,002 

,005 
,008 


31,405 
31,554 
3^705 
31,855 
32,006 
12 I C7 


33-80 
33'85 
33'90 
33^5 
34-00 


,142 
,i 4 6 
,149 
,153 
,156 


38,614 
38,786 
38,958 
39.131 
39.304 


36-05 
36-10 

36-I5 
36-20 
36-25 
76*70 


1,300 
1,303 
1,307 
1,310 

1,314 

i 718 


46,851 
47,046 
47,242 
47,438 
47,635 
47 8^2 


31-85 
31-90 

31-95 
32-00 


,014 
,018 

,021 

,024 


O^, 1 ^/ 
32,309 
32,462 
32,615 
32,768 


34-05 
34-10 

34-15 
34'20 


''59 
,163 
,166 
,170 


39,478 
39,652 
39,826 
40,002 


O u J^ 

36-35 
36-40 
36-45 
36-50 


1,321 

1,325 
1,329 
1,332 


liS* 
48,030 
48,229 
48,428 
48,627 


32-05 

32-10 

32-15 
32-20 
32-25 


,027 
,030 
,034 
,037 
,040 


32,922 
33,076 
33,231 
33,386 
33,542 
77 608 


o4 ^j 

34-30 
34-35 
34-40 
34-45 
34-50 


, J 73 
,176 
,180 
,183 
,187 
,190 


4 <J , 1 1 1 
40,354 
40,530 
40,708 
40,885 
41,064 


36-55 

36-60 

36-65 

36-70 
36-75 

16-80 


i,336 
i,340 
i,343 
i,347 
i,35i 


48,827 
49,028 
49,229 
49,431 
49,633 

AQ 8l6 


3 2 3 
32-35 
32-40 
32-45 
32-50 


43 
,047 
,050 

,053 
,056 


oj> u y 

33,855 

34,012 
34,i7o 
34,328 


34-55 
34-60 

34-65 
34-70 

3A"7C 


,194 
,197 

,201 
,204 

2O8 


41,242 
41,422 
41,602 
41,782 


36-85 

36-90 

36-95 
37-00 


,354 
1,358 
1,362 
i,365 
1,369 


4y,j 
50,039 
50,243 
50,448 
50,653 


32-55 
32-60 
32-65 
32-70 
32-75 

iv 'Xn 


,060 
,063 
,066 
,069 
,073 


34,487 
34,646 
34,8o6 
34,966 
35,126 

JC 288 


34 75 
34-8o 

34-85 
34-90 
34-95 
35-00 


,211 

,215 

,218 

,222 
,225 


4 i y u j 
42,144 
42,326 
42,509 
42,692 

42,875 


37-05 
37-10 

37-15 
37-20 

37-25 


i,373 
i,376 
1,380 


50,859 
51,065 
51,272 
5i,479 
51,687 

r T 8r>p 


3285 
32-90 
32-95 
33-00 


,<_>/U 

,079 
,082 
,086 
,089 


J3,- 600 
35,449 
35,6n 
35,774 
35,937 


35-05 
35-10 

35-I5 
35*20 


,229 
,232 
,2 3 6 

,239 


43,05 9 
43,24 4 
43,42 9 
43,6i4 
At 8no 


6/ J U 

37-35 
37-40 
37-45 
37-50 


,39* 
i,395 
i,399 
1,403 
1,406 


5 r > 5 95 

52,104 
52,3H 
52,524 
5 2 ,734 


33^5 

33-!o 

33-I5 
33-20 

33-25 


,092 
,096 

,099 
,102 

,106 


36,101 
36,265 
36,429 
36,594 
36,760 


35 2 5 
35-30 
35-35 
35'40 
35-45 
35-50 


> Z 43 

,246 
,250 

,2 3 

,257 
260 


43,500 

43.987 
44,174 
44,362 
44,550 

44,739 


37-55 
37-60 

37-65 
37-70 

3775 

27*8n 


1,410 
1,414 
1,418 
1,421 
1,425 


52,946 
53,157 

53,370 
53,583 
53,796 


33 3 
33-35 
33-40 
33-45 
33-50 


,iuy 
,112 

,116 
,119 

,122 


ju,y^o 

37,093 
37,260 

37,427 

37,595 


35-55 
35-6o 

35-65 
35-70 


,264 

,267 

,271 
,274 

2 ^C 


44,928 
45, II8 
45,308 
45,499 


67 ou 
37-85 
37-90 
37*95 
38-00 


1,429 
i,433 
i,436 
1,440 

*,444 


54,010 
54,225 
54,440 
54,656 
54,872 


33-55 
33'6o 
33-65 
33-70 
33'75 


,126 
,129 
,132 
,136 
,139 


37,764 
37,933 
38,103 

38,273 
38,443 


DJ /J 

35-8o 
35-85 
35'90 
35-95 
36-00 


j^7 
,282 
,285 
,289 
,292 
,296 


45,9 r 
45,883 
46,075 
46,268 
46,462 
46,656 


38-05 
38-10 

38-15 
38-20 
38-25 


1,448 
1,452 
i,455 
i,459 
1,463 


55,089 
55,306 
55,524 
55,743 
55,962 



444 



Appendix. 






1 


Cubes. 


Numbers. 


I 


Cubes. 


Numbers. 


ri 
1 


Cube?. 


38-30 

38-35 
38-40 

38-45 
38-50 


1,467 
i,47i 

i,475 
1,478 
1,482 


56,l82 
56,402 
56,623 
56,845 
57,067 


40-55 
40-60 
40-65 
40-70 

40-75 
40-80 
40-85 
40-90 

40-95 

4i"oo 


1,644 
1,648 
1,652 
1,656 
1,661 
1,665 
1,669 

i,673 
1,677 
1,681 


66,676 
66,923 
67,171 

67,419 
67,668 
67,917 
68,167 
68,418 
68,669 
68,921 


42'8o 
42-85 
42-90 

42'95 
43-00 


1,832 
1,836 
1,840 

1,845 
1,849 


78,403 
78,678 

78,954 
79,230 

79,507 


38-55 
38-60 

38-65 
38*70 

38-75 
38-80 
38-85 
38-90 
38-95 

39*oo 


1,486 
1,490 

1,494 
1,498 
1,502 
1,505 
i,59 
1,513 
1,517 
1,521 


57,289 
57,512 
57,736 
57,961 
58,186 
58,4H 
58,637 
58,864 
59,091 
59,319 


43*05 
43-IO 

43*15 

43*20 
43*25 
43*30 
43*35 
43*40 
43*45 
43*5o 


1,853 
1,858 
1,862 
1,866 
1,871 
1,875 
1,879 
1,884 
1,888 
1,892 


79,785 
80,063 
80,342 
80,622 
80,902 
8l,l8 3 
81,464 

81,747 
82,029 

82,313 


41*05 
41*10 
41-15 
41-20 

41*25 
41*30 

4i*35 
41-40 

4i'45 
41*50 


1,685 
1,689 

i,693 
1,697 
1,702 
1,706 
1,710 
i,7i4 
1,718 
1,722 


69,173 
69,427 
69,680 

69,935 
70,189 

70,445 
70,701 

70,958 
71,215 
71,473 


39*05 
39-10 

39*15 
39*20 

39'25 
39*30 
39*35 
39*40 
39*45 
39*50 


1,525 
1,529 
i,533 
i,537 
i,54i 
i,544 
i,548 
1,552 
i,556 
1,560 


59,547 
59,776 
60,006 
60,236 
60,467 
60,698 
60,930 
61,163 
61,396 
61,630 


43*55 
43*6o 

43*65 
43*70 

43*75 
43-80 

43*85 
43-90 
43*95 
44-00 


1,897 
1,901 

1,905 
1,910 
1,914 
1,918 

1,923 
1,927 

i,932 
i,936 


82,597 
82,882 

83,167 
83,453 
83,740 
84,028 
84,316 
84,605 

84,894 
85,184 


41*55 
41-60 

41*65 
41-70 

4i*75 
41-80 

41*85 
41-90 

4i*95 
42-00 


1,726 
i,73i 
i,735 
i,739 
i,743 
i,747 
i,75i 
i,756 
1,760 
1,764 


71,732 
71,991 
72,251 
72,512 
72,773 
73,035 
73,297 
73,560 
73,824 
74,088 


39*55 
39*6o 
39'65 
39*70 
39*75 
39*80 
39*85 
39*90 

39-95 
40-00 


1,564 
i!572 

1,576 
1,580 

1,584 

I ',592 
1,596 

1,600 


61,864 
62,099 

62,335 
62,571 
62,807 

63,045 
63,283 

63,521 
63,760 
64,000 


44*05 
44-10 

44'i5 
44-20 

44*25 
44*30 

44*35 
44-40 

44*45 
44*50 


1,940 
i,945 
1,949 
i,954 
i,958 
1,962 

i,967 
i,97i 
1,976 
1,980 


85,475 
85,766 
86,058 

86,351 
86,644 
86,938 

87,233 
87,528 
87,824 

88,121 


42*05 
42-10 

42-15 
42-20 
42-25 
42-30 

42-35 
42-40 

42-45 
42*50 


1,768 
1,772 

i,777 
i,78i 

i,785 
1,789 

1,794 
1,798 
1,802 
i, 806 


74,353 
74,6i8 
74,885 
75,i5i 
75,419 
75,687 
75,956 
76,225 

76,495 
76,766 


40-05 
40-10 

40-15 
40*20 
40-25 
40-30 

40-35 
40-40 

40-45 
40-50 


1,604 
i, 608 
i, 612 
1,616 
1,621 
1,624 
1,628 
1,632 
1,636 
1,640 


64,240 
64,481 
64,723 
64,965 
65,208 

65,45i 
65,695 
65,939 
66,184 
66,430 


44*55 
44-60 

44'65 
44-70 

44*75 
44'8o 

44-85 
44*90 
44*95 
45*oo 


1,985 
1,989 

",994 
1,998 
2,003 
2,007 
2,012 
2,016 
2,021 
2,025 


88,418 

88,717 
89,015 

89,|i5 
89,615 
89,915 
90,217 
90,519 
90,822 
91,125 


42-55 
42 '60 
42-65 
42-70 
42-75 


1,811 

1,815 
1,819 
1,823 
1,828 


77,037 
77,309 
77,58i 
77,854 
78,128 



Appendix. 



445 



1 


3 


Cubes. 


2 

1 


I 

1 


Cubes. 


d 

1 


5 


Cubes. 


3 
fc 






fc 






1 


cl 




45-05 


2,030 


91,429 


47-05 


2,214 


104,155 


49-05 


2,406 


Il8,OIO 


45' 10 


2,034 


91,734 


47-10 


2,218 


104,487 


49-10 


2,411 


118,371 


45-I5 


2,039 


92,039 


47'i5 


2,223 


104,820 


49-15 


2,4l6 


"8,733 


45-20 


2,043 


92,345 


47-20 


2,228 


105,154 


49-20 


2,421 


119,095 


45- 2 5 


2,048 


92,652 




2,233 


105,489 


49-25 


2,426 


119,459 


45-30 


2,052 


92,960 


47'3 


2,237 


105,824 


49-30 


2,430 


119,823 


45-35 


2,057 


93,268 


47-35 


2,242 


106,160 


49-35 


2,435 


120,188 


45-40 


2, 06 1 


93,577 


47*40 


2,247 


106,496 


49-40 


2,440 


120,554 


45-45 


2,066 


93,886 


47'45 


2,252 


106,834 


49-45 


2,445 


120,920 


45-50 


2,070 


94,196 


47-50 


2,256 


107,172 


49-50 


2,45 


121,287 


45-55 
45-6o 


2,075 
2,079 


94,507 
94,8i9 


47-55 
47-60 


2,261 
2,266 


107,511 
107,850 


49-55 
49-60 


2,455 
2,460 


121,655 
122,024 


45-65 


2,084 


95,i3i 


47-65 


2,271 


108,190 


49-65 


2,465 


122,393 


45-70 


2,088 


95,444 


47-70 


2,275 


108,531 


49-70 


2,470 


122,763 


45-75 


2,093 


95,758 


4775 


2,280 


108,873 


4975 


2,475 


123,134 


45-80 


2,098 


96,072 


47-80 


2,285 


109,215 


49-80 


2,480 


123,506 




2, IO2 


96,387 


47*85 


2,290 


109,558 




2,485 


123,878 


45-90 


2,107 


96,703 


47-90 


2,294 


109,902 


49-90 


2,490 


124,251 


45-95 


2,111 


97,019 


47-95 


2,299 


110,247 


49-95 


2,495 


124,625 


46*00 


2,116 


97,336 


48-00 


2,304 


110,592 


50-00 


2,500 


125,000 


46-05 


2,121 


97,654 


48-05 


2,309 


110,938 








46-10 


2,125 


97,972 


48-10 




111,285 




46-15 


2,130 


98,291 


48-15 


2^318 


111,632 




46-20 


2,134 


98,611 


48-20 


2,323 


III,98o 




46-25 


2,139 


98,932 


48-25 


2,328 


112,329 




46-30 


2,144 


99,253 


48-30 


2,333 


112,679 






2,I 4 8 


99,575 


48-35 


2,338 


113,029 




46-40 


2,153 


99,897 


48-40 


2,343 


"3,380 




46-45 


2,158 


100,221 


48*45 


2,347 


"3,732 




46-50 


2,l62 


100,545 


48-50 


2,352 


114,084 




46-55 


2,l67 


100,869 


48-55 


2,357 


"4,437 




46-60 


2,172 


101,195 


48-60 


2,362 


"4.79 1 




46-65 


2,176 


101,521 


48-65 


2,367 


115,146 




46-70 


2,181 


101,848 


48-70 


2,372 






46*75 
46-80 


2,186 
2,190 


102,175 
102,503 


48-75 
48-80 


2,377 
2,381 


"5^57 
116,214 




46-85 


2,195 


102,832 


48-85 


2,386 


116,572 




46-90 


2,200 


103,162 


48-90 




116,930 




46-95 


2,204 


103,492 


48-95 


2^396 


117,289 




47-00 


2,209 


103,823 


49-00 


2,401 


117,649 





APPENDIX C 

SYLLABUS OF EXAMINATIONS IN SUBJECT u. 
NAVAL ARCHITECTURE 

The students should be encouraged to make good rough 
sketches of the different parts of a ship's structure approximately 
to scale, using squared paper ; they should also be impressed with 
the necessity of noting any detail of work brought before their 
notice daily in the shipyard. Questions will be set in the exami- 
nation which require rough sketches of parts of a vessel to be 
given from memory. 

If the class is held in an institution which possesses a testing 
machine, the students ought to be allowed to use it occasionally to 
test samples of materials used in shipbuilding. 

All students should be provided with suitable scales, set 
squares, and ship curves, and candidates should bring these 
to the examination. 

Table of logarithms, functions of angles, and useful constants 
will be provided, and candidates will be restricted to use of these 
tables, and will not be allowed to bring with them into the 
examination room any other mathematical or logarithm tables. 
Slide rules may be used. 

Compulsory questions may be set in either of the examination 
papers. 

LOWER EXAMINATION 

I. PRACTICAL SHIPBUILDING. The tests to which the various 
materials used in shipbuilding are subjected, and the defects to 
which those materials are liable ; the tools and appliances used in 
ordinary shipyard work, and the general arrangement of blocks, 



Appendix. 447 

staging, derricks, etc., used on a building slip ; plans of flat and 
vertical keels, inner bottom, shell, deck and other plating; 
framing, beam, keelson, and stringer plans ; watertight and other 
bulkheads ; ceiling and wood decks ; pillaring arrangements to 
secure clear holds, and details of cargo hatchways to meet Lloyd's 
Rules ; rudders, stern frames, and spectacle arrangements for 
twin-screw ships ; bilge keels ; supports to engines, boilers, and 
shafting ; masts and derricks ; precautions necessary to prevent 
deterioration of the hull of a ship while building, and while on 
service ; method of docking ships, how they are placed in position 
and supported. 

II. LAYING OFF. A knowledge of the work carried on in the 
Mould Loft for the purpose of fairing a set of lines, including 
traces of keelsons and longitudinals, edges of shell plating, tank 
margins, ribbands, etc., and transferring the frame and other lines 
to the scrive board ; lifting the bevels and constructing round of 
beam mould ; a ship's block model and the information necessary 
for its construction ; obtaining the dimensions for ordering the 
shell plating, frames, beams, floors, inner bottom plating, etc. ; 
making and marking ribbands ; fairing the edges of shell plating 
on the frames ; making templates or skeleton patterns for stem, 
sternpost, propeller bracket forgings or castings. 

III. DRAWING. Plotting of curves of displacement, tons per 
inch immersion, I.H.P., etc., from given data. A rough freehand 
dimensioned sketch may be given at the examination, requiring 
candidates to make finished scale drawings, and candidates will 
be expected to be able to draw, from their own knowledge, the 
fastenings suitable for connecting together the parts which are the 
subject of the example. 

IV. SHIP CALCULATIONS. Calculation of the weights of 
simple parts of a ship's structure ; spacing and strength of iron 
and steel rivets ; calculation of the strength of the simple parts 
of a ship's structure, such as tie plates, butt straps and laps ; tons 
per inch immersion ; change of trim, and moment to change trim; 
change of trim due to moving weights on board, and that due to 
the addition or removal of weights ; the principles and use of 
Simpson's and other rules for finding the area and position of the 
centre of gravity of a plane area, and for calculating the position of 
the centre of buoyancy ; graphic methods of finding displacement 
and position of the centre of buoyancy ; curves of displacement 
and of tons per inch immersion ; the fundamental conditions to be 
fulfilled in order that any body may float freely and at rest in still 
water ; centre of flotation, metacentre, metacentric height, stable 



448 Appendix. 

and unstable equilibrium ; definitions of block, prismatic, water- 
plane, midship area, and other similar coefficients. 

The questions will be of the same type as those set in Stage 2 
of previous examinations. 

HIGHER EXAMINATION 

I. PRACTICAL SHIPBUILDING. The structural arrangements 
necessary to resist longitudinal and transverse stresses to which 
ships are liable in still water and amongst waves, and the arrange- 
ments to resist local stresses ; description and rough hand sketches 
of detail fittings of ships, such as anchor and capstan gear, steering 
gear, and other appliances used in working a ship ; davits and 
fittings in connection therewith ; ventilating and coaling arrange- 
ments ; pumping and draining ; the fundamental types of vessels 
and modifications thereto, the distinctive features of such vessels 
and consequent effect on freeboard ; methods of determining the 
sizes of structural parts and of detail fittings making out midship 
sections to the Rules of the principal classification Societies for 
various types of vessels ; methods of fitting up refrigerating spaces 
for shipment of frozen and chilled meat, fruit, etc. ; construction of 
oil fuel bunkers, and of vessels for carrying oil ; launching arrange- 
ments, and the diagrams and curves generally used in connection 
therewith. 

LAYING OFF. Expanding the plating of longitudinals and 
margin plates by the geometric and mocking up methods ; ex- 
panding stern plating, rudder trunking, and mast plating ; 
obtaining the true shape of a hawse hole in the deck or shell, and 
similar practical problems ; constructing and fairing the form 
of a twin screw bossing. 

III. SHIP CALCULATIONS. Displacement sheet and arrange- 
ment of calculations made thereon ; proofs of Simpson's and other 
rules for obtaining areas and moments ; displacement and dead- 
weight scales ; approximate and detailed calculations relating to 
the weight and position of the centre of gravity of hull ; calculations 
of weight and strength of parts of a ship's structure such as decks, 
bulkheads, framing, side and bottom plating, etc., also the strength 
of fittings such as boat davits, derricks, etc. ; coefficients of weight 
of hull, outfit, and machinery for a few of the principal types of 
ships, also coefficients of position of the centre of gravity of the 
ships ; curves of loads, shearing forces, and bending moments for 
a ship floating in still water, and amongst waves, also equivalent 
girder and stress in the material ; calculations of the positions 



Appendix. 449 

of transverse and longitudinal metacentres ; consideration of the 
curves of centres of buoyancy, centres of flotation, and pro-meta- 
centres; the construction and use of metacentric diagrams; 
Attwood's and Moseley's formulae, and methods of calculating 
stability based thereon ; the construction and use of curves of 
stability ; inclining experiment and the precautions that must 
be taken to ensure accuracy ; change of draught and trim due 
to passing from fresh into salt water and vice versa ; effect upon 
trim and stability due to flooding compartments of a ship ; effect 
of free surface on the stability of vessels carrying liquid cargo ; 
methods of determining the size of rudder-heads, and the stresses 
on rudders balanced and unbalanced ; resistance of ships ; 
Froude's experiments on skin friction ; Froude's law of comparison 
for vessels at corresponding speeds ; methods of calculating the 
horse-power to propel a vessel of known form at a given speed ; 
effective horse-power, propulsive coefficient and Admiralty con- 
stants, and values of the two last in typical cases ; speed of ships 
on trial, methods adopted and precautions necessary to obtain 
accurate speed data ; progressive trials and their uses ; elementary 
considerations of the oscillations of ships in still water and 
amongst waves; definitions of a "stiff" and "steady" vessel, 
and elements of design affecting these qualities ; tonnage of ships, 
how measured, etc. 

The questions in this stage will correspond generally in type 
with those set in the Stage 3 examination held under the previous 
regulations, but the standard will be higher, both for a Pass and 
a First Class. 



2 



APPENDIX D. 

1902. 

ELEMENTARY STAGE. 

General Instructions. 

You are permitted to answer only eleven questions. 

You must attempt No. II. Three of the remaining questions 
should be selected from the Calculations \ and the rest from the 
Practical Shipbuilding section. 

PRACTICAL SHIPBUILDING. 

In Questions I to 9 inclusive your answers may be given in 
reference to any type of ship. The type selected should be named 
in each question, and the scantlings given in each case. 

1. Make a rough sketch showing the cross-section, on a scale 
of about ^ the full size, of the flat keel plate and its connections to 
the vertical keel plate and to the garboard strake. (6) 

2. What is the spacing of the rivets in the edges of the keel 
plate and the garboard strake, and in the butt straps of the vertical 
keel plate ? (8) 

3. Make a sketch showing, for a transverse frame, a cross- 
section (a) through a floor, and (b) through a frame above the 
floor. (8) 

4. Make a sketch showing the section of a stem connection to 
ordinary outer bottom plating near the water-line. (6) 

5. Make a sketch of the cross-section of the arms of an A 
bracket or strut of a twin-screw ship, giving dimensions. (6) 

6. Make a sketch showing a horizontal section through a 
transverse water-tight bulkhead. (8) 

7. Make a transverse section through the sheerstrake and upper 
deck stringer, showing the rivet connection of these plates to the 
adjacent strakes and to each other ; also the beam and its rivet 
connection to the frames. (8) 



Appendix. 45 1 

8. Sketch one method of constructing the heads and heels of 
pillars, showing the rivet connections. (8) 

9. Sketch a butt strap connection of outer bottom plating ; 
show the spacing and size of rivets in it. (8) 

10. Sketch a disposition of butts in the bottom planking of a 
sheathed ship, giving specimens of the positions and sizes of the 
fastenings. (10) 

DRAWING. 

11. Enlarge sketch No. n to a scale of twice that upon which 
it is drawn. (This sketch was a portion of the after body of a twin- 
screw ship, showing frame lines in way of shaft.) (16) 

CALCULATIONS. 

12. What is the relation which must exist between the weight 
of a body floating freely at rest in a liquid and the volume of its 
submerged part ? A body of uniform circular transverse section 
floats freely in sea-water so that the centres of the circular sections 
are in the water surface. What will its weight be if its length is 
100 feet and its diameter 20 feet ? (8) 

13. Find the area of a half of a ship's water-plane of which the 
curved form is defined by the following equidistant ordinates 
spaced 12 feet apart : 

o-i, 5-1, 7-17, 875, io-i, 9-17, 8-05, 6-4, o-i feet. (6) 

14. By what number would you have to divide the area in 
square feet of a water-plane in order to obtain the number of tons 
weight it would be necessary to add to the ship in order to increase 
her draught one inch in salt water ? (6) 

15. What is the relative position of the centre of gravity of the 
weight of a body floating freely at rest in water and the centre of 
gravity of the volume of the submerged portion of the body ! What 
is the condition necessary for stable equilibrium ? (8) 

1 6. What are the weights of a cubic foot of steel, yellow pine, 
and copper? What is the weight of a hollow steel pillar 10 feet 
long whose external diameter is 5 inches and internal diameter 
4 inches ? What is the diameter of a solid pillar of the same 
weight? (10) 



452 Appendix. 



ADVANCED STAGE. 
Instructions. 

You are permitted to answer only twelve questions. 

You must attempt Nos. 22 and 28. The remaining questions 
may be selected from any part of the paper in this stage, provided 
that one or more be taken from each section, viz. Practical Ship- 
building, Laying Off, and Calculations. 

PRACTICAL SHIPBUILDING. 

Questions 17 to 22, inclusive, may be answered with reference 
to any one type of ship to which the question may apply, but each 
type referred to must be named in the question in which it is dealt 
with. 

17. Where is the material of a ship's structure most severely 
stressed, and under what conditions ? (10) 

1 8. Give a sketch showing the disposition and size of rivets in 
a buttstrap connection of (i) a sheerstrake, (2) a weather-deck 
stringer bar. Sketch a bulkhead liner. (13) 

19. Sketch the transverse framing in a double bottom, giving 
the scantlings and the disposition and sizes of the rivets. (14) 

20. Describe the operation of framing a ship from the beginning 
of handling the unmarked plates and bars to the time the framing 
is faired. (20) 

21. Sketch the blocks upon which a ship is built, giving the 
spacing and sizes of the blocks. (10) 

22. Sketch a rudder, giving sizes ; also give the sizes and dis- 
position of rivets, pintles, and bolts. (13) 

23. What is the breaking stress and elongation per cent, in 8 
inches of mild steel ? Give the same with reference to any high- 
tension steel, and name the class of vessel in which it is used. 

(10) 



Appendix. 453 

24. What is the relative elasticity of yellow pine and steel in 
combination in a deck in compression and extension ? Sketch an 
ordinary disposition of butts and bolts in a wood deck, estimating 
the effective area in compression and also in extension. (13) 



LAYING OFF. 

25. What information is given to the mould loft to enable the 
loftman to lay down the lines of a ship ? (10) 

26. Show how to obtain the development and the projection in 
the sheer and half-breadth plans of a diagonal. Explain how to 
obtain the ending at the stem in the developed diagonal. (18) 

27. Show how to find the point where a straight line not parallel 
to any of the planes of projection (the sheer, half-breadth, or body 
plan) would cut the surface of the ship. (20) 



DRAWING. 

28. What does Sketch No. 28 represent ? Enlarge it to twice 
the scale upon which it is drawn. (This sketch was the sternpost 
and rudder of a screw ship in profile.) (25) 



CALCULATIONS. 

29. Calculate the volume and position of centre of gravity, 
horizontally and vertically, of a form given by the following 
ordinates : 

ft. ft ft. 

No. i W.L. ... o'i 7*17 io - i 8*05 o'i 

No. 2 ... o'i 5'66 8-0 6*46 o'i 

No. 3 ... o'i o'i o'i o'i o-i 

Horizontal interval, 24 feet ; vertical interval, 3 feet. 

(18) 

30. The areas of transverse vertical sections of a solid are 1*2, 
61-2, 86'o, 121-0, 96-6, 76-8, 1-2 square feet, at distances apart of 
12, 12, 24, 24, 12, and 12 feet, respectively. Find the volume and 
longitudinal position of the centre of gravity of the solid. (10) 



454 Appendix. 

31. What is the transverse metacentre of a ship in the upright 
position ? What is the value of the distance between this meta- 
centre and the centre of buoyancy ? Find the value of this distance 
for the largest water-line given in Question 33 (assuming a value 
for the displacement and position of centre of buoyancy if Question 
33 has not been attempted). (18) 

32. A hold beam is formed of two beams, each formed of a 
-inch plate 12 inches deep, and four angles 4" x 4" x " I the 
beams are connected together by a top plate inch thick, extending 
from the fore edge of the flange of the forward beam to the after 
edge of the flange of the after beam. Find the weight of 30 feet of 
such a beam. The frame spacing is 24 inches. (18) 



HONOURS. PART I. 
Instructions. 

You are premitted to answer only fourteen questions. You 
must attempt Nos. 43 and 48 ; the remainder you may select from 
any part of the paper in this stage, provided that one or more be 
taken from each section, viz. Practical Shipbuilding, Laying Off, 
and Calculations. 

PRACTICAL SHIPBUILDING. 

Questions 33 to 42 may be answered with reference to any 
suitable type of ship, but the type must be mentioned, and the 
principal figured dimensions inserted. 

33 Make a sketch of a stern-frame of a twin-screw vessel. 

(20) 

34. Make a sketch of the method of making the following water- 
tight in passing through a bulkhead : 

(i) A bulb tee ; (2) a keelson formed of a plate and four angle- 
bars. (13) 



Appendix. 455 

35. Make a sketch of a boiler hatch-coaming, showing the con- 
nection to the half ^ earns and to the casing-plates. Show what 
method is adopted to strengthen the deck at the corners of the 
hatch. (15) 

36. Make a sketch of a right and left-handed screw steering- 
gear, giving the sizes and materials of the different parts. Show 
the method of working by hand as well as steam. (21) 

37. Sketch the stowage of an anchor, showing the position of 
the leads to the capstan and cathead. (21) 

38. Sketch the arrangements made for the launch of a ship. 
State the declivity and camber of the launching ways, and the 
declivity of the keel. (21) 

39. Describe the pickling process for steel plates. What pro- 
tective materials are put upon the various parts of a steel ship ? 

(16) 

40. Describe briefly the steam pumping and drainage arrange- 
ments of a ship, detailing where each steam-pump draws from and 
delivers to. (21) 

41. Make a sketch showing a section of a steel mast, and show 
a disposition of butts of plates, with arrangements of rivets in the 
buttstrap. (16) 

42. Make a sketch of a pair of davits, giving the cleats and 
blocks attached to it. Show the method of securing the boat in- 
board, and state how the position of the davits is fixed in relation 
to the boat. (21) 

LAYING OFF. 

43. Make a sketch of a body plan on a scrive-board, showing 
all the lines that are put upon it, stating what each line is for. 

(22) 

44. Show how you would obtain the exact, form of the projection 
of the intersection of a large conical pipe, with the outer bottom- 
plating of a ship, the axis of the. cone not being parallel to either 
the sheer, half-breadth, or body plan. (21) 

45. Given the body plan of a ship without any bossing out in 
way of the shaft, describe how you would obtain the form of this 
bossing in the case of a twin-screw ship. (21) 



456 



Appendix. 



CALCULATIONS. 

46. Find the displacement up to the 6-feet and lo-feet water- 
lines of a ship whose form is defined by the following : 





W.L.'s. 


Keel. 


I ft. 


2ft. 


4 ft. 


6ft. 


8ft. 


10 ft. 

I 


Nos. of Sec 

Distance 
between 
sections f 
is 
40 feet. 


tion. 

/i 

i* 

2 

3 
4 
5 
6 

6* 

\7 


O'l 


O'l 


OT 


O'l 


O'l 


O'l 


O'l 


i '4 


2'6 


4*6 


67 


9-0 


II'I 


O'l 


5'6 


8'2 


"'5 


IS-S 


H7 


I5'3 


O'l 


ii'i 


137 


15-9 


16-7 


17-0 


i6'9 


O'l 


13-1 


I 5 '6 


17-1 


17-1 


17-4 


17-4 


O'l 


10-3 


I2'6 


14-6 


I5-4 


15-8 


16-0 


O'l 


57 


7-5 


9*5 


107 


ii'6 


12-5 


O'l 


17 


27 


4'i 


5'o 


6-0 


9-1 


O'l 


O'l 


O'l 


O'l 


O'l 


O'l 


O'l 



(21) 

47. Find the vertical and longitudinal position of the centre of 
buoyancy of the form in the preceding question. (18) 

48. Suppose the vessel in Question 54 to be floating at the loft, 
water-line, and to be inclined transversely through an angle of one 
in one hundred, by a weight of one ton moved through 30 feet. 
Find the height of the centre of gravity of the ship above the keel. 

(If Questions 46 and 47 have not been done, assume a displace- 
ment and height of C.B.) (18) 

49. What is the ultimate shearing and tensile stresses of steel 
rivets and plates respectively? Find the breaking strength of a 
single butt, double-chain riveted, of a plate 30 inches wide, inch 
thick, connected by rivets, spaced 3 inches apart. Find the force 
necessary to break the plate across a frame-line where the rivets 
are spaced 6 inches apart. (21) 

50. Given the height of the longitudinal metacentre above the 
centre of gravity, show how you would obtain the moment to trim 
ship one inch. In a vessel whose distance between draft marks 
forward and aft is 300 feet, and whose centre of gravity of water- 
plane is 10 feet abaft centre between draft marks, and whose foot- 



Appendix. 457 

tons to trim ship one inch are 400, find the change of draft forward 
and aft caused by moving 30 tons through 200 feet in a fore and 
aft direction. (21) 

51. A barge 100' x 20' x 10' of rectangular section, is formed 
of -inch plating on ends, bottom, sides, and deck, and has frames 
and beams of 4^" x 4" x ", spaced 20 inches apart. The ends 
have stiffeners 2 feet apart, of the scantlings of the frames. A floor- 
plate, 12" x ", is on every frame. Find the weight of the hull, 
assuming that there are no hatches. Suppose the barge to have 
weights of 10 tons at 10 feet from the stem, 15 tons at 25 feet, 20 
tons at 50 feet, 30 tons at 75 feet. Find the longitudinal position 
of the centre of gravity of the loaded barge. (30) 



HONOURS. PART II. 

Instructions. 

You are not permitted to answer more ihanfvurteen questions. 

NOTE. No Candidate will be credited with a success in Part 
II. of Honours who has not obtained a previous success in Honours 
of the same subject. 

Those students who answer the present paper sufficiently well 
to give them a reasonable chance of being classed in Honours, 
will be required to take a practical examination at South Kensing- 
ton. Honours Candidates admissible to this Examination will be 
so informed in due course. 

52. State and prove Simpson's First Rule. State Tchebycheffs 
Rules for either three, five, or seven ordinates. (25) 

53. Prove that in a curve of loads of a ship floating at rest, the 
integral of the area of the curve from one end up to a chosen point 
gives the shearing force at that point, and that the integral of the 
curve of shearing forces over the same part gives the bending at 
the chosen point. (25) 

54. Suppose a curve of buoyancy to be a curve of versed sines, 
and the corresponding curve of weights to be a common parabola, 
whose axis is vertical and at the middle of the length. Find the 
form of the curves of shearing force and bending moment. (33) 

55. State and prove Atwood's formula. (21) 

56. Describe any method of obtaining a cross-curve of stability. 

(25) 



45 8 Appendix. 

57. Find the effect upon the draught of water forward and aft 
of opening to the sea one per cent, of the length of a vessel of 
rectangular section at any part of the length, supposing this one 
per cent, to be confined between transverse water tight bulkheads. 
Explain how the deductions from this result can be made use of 
to determine the spacing between water tight bulkheads, which 
shall not be exceeded, in order that when a compartment is flooded, 
the draft in no case shall exceed a certain specified amount. (33) 

,58. If B and B l be respectively CB's in upright and inclined 
position, and R be the foot of the perpendicular from B on to the 
vertical through B^ in this inclined position, show that B\R is the 
integral of BR between the upright and the angle of inclination. 
Show from this how a curve of CB's can be obtained from a curve 
of GZ's. (25) 

59. A vessel of uniform rectangular section is launched parallel 
to her keel. What is the form of the curve of tipping and lifting 
moments, supposing the ship to be deep enough to prevent the 
upper deck from being immersed ? Suppose such a vessel 100 feet 
long, 24 feet wide, having its CG at the middle of the length to be 
launched at a slope of one inch to the foot, with two feet of salt 
water over the end of the ways, and with a launching weight of 100 
tons. What is the maximum pressure on the fore end of the ways 
(assumed to be at the fore perpendicular) ? (43) 

60. Suppose a vessel to be instantaneously floating on the crest 
of a wave of her own length, and of height equal to one-twentieth of 
the length. What stress would you expect to find with all coal 
burnt out in 

(1) a battleship of 14,000 tons ; 

(2) a high-speed Atlantic liner ; 

(3) a torpedo-boat destroyer ? 

Give figures for the stress when in the hollow of the same wave with 
bunkers full. (25) 

61. Why would a trochoidal wave cause less stress than that 
determined in the preceding question? Give results of any 
calculations you know of, in which the difference due to the wave 
not being actually at rest is taken into account. (25) 

62. Prove that in a ship, whose moment of inertia about a 
transverse axis through the midship section is the same for the 
fore end as for the after end, the pitching does not alter the bend- 
ing moment at the midship section. (25) 

63. Find the maximum stress upon a section of a vessel floating 
upright in still water, and subjected to a bending moment of 1000 



Appendix. 

foot-tons. The section is rectangular, 20 feet wide, 10 feet deep, 
and has " plating on deck, bottom, and sides. 

Suppose the vessel to be inclined at some known angle, how 
would you find the maximum stress ? (33) 

64. What is the Admiralty speed coefficient ? What is its value 
for the vessels named in Question 60 at full speed? How does it 
vary with speed? What is its value for sea work as compared 
with trial trips ? (33) 

65. A model 12 feet long, having a displacement of 1000 Ibs., 
has a resistance of 3 Ibs. at 5 feet per second. Find the effective 
horse-power necessary to drive a vessel of the same form 192 feet 
long at its corresponding speed. Assume the wetted surface of the 
model to be 30 square feet, and the frictional resistance of a plane 
12 feet long at 5 feet per second to be 0*07 Ib. per square foot, and 
that of a plane 192 feet long to be o'8 Ib. per square foot at 12 
knots. (25) 

66. What conditions should be fulfilled in a ship to make her 
easy in her rolling at sea ? (25) 

67. What is a curve of extinction? How can it be obtained 
experimentally? What can be determined from it in relation to 
the resistance to rolling of a ship ? (25) 

68. What is the chief cause of vibration in a steamer ? What 
are the subsidiary causes ? What precautions are taken to avoid 
these ? (25) 



1905. 
STAGE II. 

Instructions. 

You are permitted to answer only eight questions. 

You must attempt Nos. 32 and 35. The remaining questions 
may be selected from any part of the paper in this stage, provided 
that one or more be taken from each section, viz. Practical Ship- 
building, Laying Off, and Calculations. 

PRACTICAL SHIPBUILDING. 

21. Sketch a good disposition of butts for outer bottom plating, 
assuming your own spacing of frames and length of bottom plates, 



460 Appendix. 

stating the spacing of frames and length of plates you have 
assumed. (20) 

22. For what purposes are " liners " and " bulkhead liners " 
fitted in ships ? Describe a system of construction in which 
ordinary liners are not necessary. Sketch a bulkhead liner. (20) 

23. Sketch and briefly describe any efficient type of steering 
gear. (20) 

24. State what parts of a vessel are most effective in resisting 
longitudinal stresses, and give the reasons for your answer. 

(25) 

25. Sketch and briefly describe the construction of a steel mast, 
stating the size of mast, scantlings, and size and pitch of rivets. 

(20) 

26. Show by sketches how a modern hawse pipe is secured to 
the structure of a ship. Of what materials are hawse pipes made, 
and how is the diameter determined ? (20) 

27. How is a large transverse watertight bulkhead plated, stiff- 
ened, and secured ? (25) 

28. State the tests made to ensure that either large or small 
steel castings, and steel plates are fit for use in shipwork. Describe 
how the mill-scale formed on steel plates during manufacture can 
be removed. (20) 

LAYING OFF. 

29. Briefly describe the contracted method of fairing the body. 
If the floor be of small length, what precautions are necessary when 
fairing a long ship ? (20) 

30. How would you obtain the lines of the inner bottom on the 
floor, and fair them ? How would you arrange and fair the plate 
edges ? (20) 

31. Describe how an account is obtained of the outside plating 
of a steel ship in order that the plates may be demanded from the 
manufacturer. What margin would you allow at the edges and 
butts ? (25) 

DRAWING. 

32. What does the given sketch represent? It is drawn on a 
scale of " to i foot ; draw it in pencil on a scale of " to I foot. 

(35) 
CALCULATIONS. 

33. The tons per inch immersion of a ship at seven equidistant 
water-lines 3 feet apart are respectively 31-3, 30-2, 28-5, 26-4, 23 9, 
19-6, and 14-2. Find the displacement and the vertical position of 



Appendix, 4^ T 

the centre of buoyancy. The appendage below the lowest water- 
plane to be neglected. (22) 

34. What conditions have to be fulfilled in order that any body 
may float freely and at rest in still water ? What is the condition 
necessary for stable equilibrium ? (20) 

35. State Simpson's second rule. 

The equidistant half-ordinates of the load water-plane of a ship 
in feet are o'6, 2*9, 9-1, 15*6, i8'o, 187, 18-5, 17-6, 15-2, and 67 
respectively, and the length of the ship is 288 feet. Find the area 
of the load water-plane and the longitudinal position of the centre 
of gravity. (25) 

36. Describe the process known as the "graphical process," 
used for finding the displacement and centre of buoyancy of a ship. 

(23) 



STAGE III. 
Instructions. 

You are permitted to answer only eight questions. You must 
attempt No. 51, and one other question at least should be selected 
from the Calculations. 

PRACTICAL SHIPBUILDING. 

41. To what stresses is the hull structure of a ship subjected ? 
Describe how they are set up, and the provision made to meet 
them. (35) 

42. Describe briefly the principal pumping and draining arrange- 
ments of a large ship, naming the type of ship selected. (35) 

43. Roughly sketch the midship section of- a vessel, giving 
scantlings. Name the type of vessel selected. (35) 

44. Sketch and describe the usual methods of constructing and 
fitting watertight sliding doors. How are they opened and closed ? 

(35) 

45. Describe, with sketches, the construction and uses of the 
following shipyard machines, viz. : 

(a) Plate bending rolls. 
(b} Punching and shearing machine. 

Show in detail the construction of the latter machine in way oi 
the punch. (35) 

46. Sketch and describe the special features of the general 
arrangement and detail construction of the hull of an oil steamer. 

(35) 



462 



Appendix. 



47. Sketch and describe the construction of a bridge. How is 
it supported and stiffened against rolling strains ? Enumerate the 
fittings generally placed on the bridge, and show how they are 
arranged. (35) 

48. Describe how the scantlings of a ship of known type and 
dimensions are determined by Lloyd's Rules. What portions of 
the hull structure are determined by the ist and 2nd numbers 
respectively ? (40) 

LAYING OFF. 

49. Describe how the line of centre of shaft is got in on the 
floor, and the body faired into the shaft tube of a twin screw ship. 

(35) 

50. How would you obtain a correct mould for a longitudinal, 
or tank margin plate, which has considerable twist and curvature ? 

What marks would be placed on the mould for the information 
and guidance of the workman ? (35) 



CALCULATIONS. 

51. Calculate the displacement and vertical position of the 
centre of buoyancy of a vessel for which the half-ordinates are 
given below, the distance between the sections being 14 feet, 
and the keel appendage being 2'6 tons, with centre of buoyancy 
4-8 feet below the 5' 6" water-line. 

(45) 



Sections. 


,W.L. 


i' 9" W.L. 


a' 6" W.L. 


4 ' W.L. 


5 '6"W.L. 


! 


O'l 


o-3 


0-8 




2'8 


2 


2'2 


3-9 


5'4 


6-6 


7* 


3 


3'4 




6'8 


7*4 


7*5 


4 


2*O 


3*6 


5' 1 


6-4 


6-8 


5 


O'l 


0'2 




1*4 


2-8 



52. Define change of trim) and moment to change trim one inch. 
Obtain an expression for the position in which a weight must 

be placed on board a ship so as not to increase her maximum 
draught. Explain clearly why this is not always possible with 
large weights, and find the limiting weight. (35) 

53. What are the curves of displacement and tons per inch 
immersion, and what are their uses ? 



Appendix. 463 

The areas of a ship's sections at parallel water-lines 3 feet apart 
are 9600, 9500, 9000, 7700, 5000, and 2000 square feet. Neglecting 
the volume below the lowest section, find the tons per inch at each 
water-plane, and plot the curve of tons per inch. Find also the 
total displacement. (35) 

54. What are the ultimate shearing and tensile stresses of steel 
rivets and plates respectively ? 

Two tie plates 24" wide by f" thick, are connected together by 
a lapped joint. Show by calculation the number and sizes of rivets 
required, indicating how they should be arranged in order that the 
butt and plate may be nearly of equal strength. (35) 

55. Define "centre of flotation," " centre of buoyancy," "meta- 
centre," and " metacentric height." 

Determine the distance between the centre of buoyancy and 
the transverse metacentre of a vessel 72 feet long and 95 tons 
displacement, floating at a water-plane whose half-ordinates are, 
0-8, 3'3, 5*4, 6-5, 6-8, 6-3, 5-1, 2-8, and 0-5. (40) 

56. A fore-and aft watertight bulkhead, extending from the 
tank top to main deck, is 50 feet long and 24 feet deep. Find the 
total weight of the bulkhead, including stiffeners, connecting angles, 
etc., having given the following particulars : 

Plating jV thick for the lower half depth, and J" above, with 
single-riveted edges and butts ; stiffeners alternately 6" x 3" x 3^" 
zed bars of 15 Ibs. per foot run, and 3^" x 2^" angle bars of 7 Ibs. 
per foot run, spaced 2 feet apart ; bounding angles 3^" x 3" of 
8'5 Ibs. per foot run. (35) 



HONOURS. 

Instructions. 

You are not permitted to answer more than eight questions. 

NOTE. No Candidate will be credited with a success in this 
examination who has not obtained a previous success in Stage II 7. 
or in Honours, of the same subject. 

61. State and prove Simpson's second rule for approximating 
to the area and centre of gravity of a plane surface. 

The half-ordinates of a water-plane are 0*2, i'8, 4'8, 7*4, 5*5, 2*3, 
and 0*6 feet. The ordinates are spaced 23 feet apart. Find the 
distance of the centre of gravity of the half water-plane from the 
middle line. (40) 



464 Appendix. 

62. Obtain an expression for the height of the metacentre above 
the centre of buoyancy in a floating body. 

The half ordinates of the water-plane of a vessel, 27^ feet apart 
are o'i, 6*9, io'o, 10-5, 10*1, 7*2, and o'i feet respectively. Deter- 
mine the transverse metacentric height, having given that the dis- 
placement to the water- plane (salt water) is 275 tons, and that the 
centre of gravity of the vessel is 5 feet above the centre of buoyancy. 

State the values of the metacentric heights in any two types 
of ships with which you are acquainted, naming the types of vessels 
selected. (50) 

63. The maximum speed of a vessel is 17 knots, and the rudder, 
which is 12 feet broad and approximately rectangular in shape, has 
an area of 200 square feet and a maximum working angle of 
35 degrees. Estimate the diameter required for the rudder head 
if made of cast steel. 

How does the case of a balanced rudder differ from that of an 
ordinary one in the case (a) when the ship is going ahead, and (b} 
when she is going astern ? (45) 

64. What are " cross curves of stability " ? 

Describe fully how you would construct a set of cross curves of 
stability for a vessel of known form. Explain clearly the great 
advantages of having stability calculations recorded in this form. 

(45) 

65. Show how you would estimate the angle of heel to which 
a ship under sail in still water would be driven, when struck by a 
squall of known force, (a) when the ship is upright and at rest ; and 
() when the ship has just completed a roll to windward, when the 
squall strikes her. (45) 

66. Prove that in a curve of loads of a ship floating at rest, the 
integral of the area of the curve from one end up to a chosen point 
gives the shearing force at that point. 

At the section of a ship at which the shearing force is at a 
maximum, show how the shearing stress on the material varies, 
and state under what circumstances this shearing stress would cause 
straining action to take place. (50) 

67. Describe the principles governing the watertight subdivision 
of war or merchant ships. Is there any legal enforcement for 
merchant ships ? 

State briefly the recommendations of the Bulkheads Committee 
(1890-91). 

A barge is of uniform rectangular section, 70 feet long and 
20 feet broad, and the draught of water when the vessel is 
intact is 8 feet. What would be the minimum height of a bulkhead 



Appendix. 465 

lo feet from one end of the vessel in order that if the end compart- 
ment were flooded, the adjacent compartment should remain dry ? 

(50) 

68. Define Statical Stability and Dynamical Stability. 

A submarine vessel 140 feet long has a uniform cross section 
of which the upper part is a semicircle 10 feet in diameter, and the 
lower a triangle 8 feet deep with vertex downwards. The centre of 
gravity of the vessel is 6 feet above the keel. 

Construct, to scale, the curve of statical stability, and state 
in foot-tons the dynamical stability at 60 degrees. (50) 

69. State what is meant by " effective horse-power," " propulsive 
coefficient," and " corresponding speeds." State the values of the 
propulsive coefficients of any two types of vessels with which you 
are acquainted, naming the types selected. How does the pro- 
propulsive coefficient vary with the speed in a particular ship, and 
why? 

A vessel of 1800 tons displacement is propelled at 15 knots by 
engines of 2500 I.H.P. Estimate the I.H.P. you would consider 
necessary to drive a vessel of similar model, but of 4000 tons dis- 
placement at a speed of 18 knots. What assumptions are made in 
passing from the one vessel to the other ? (45) 

70. Describe briefly the causes which produce vibration in the 
hulls of steamships, and state under what circumstances these 
vibrations reach a maximum. 

(a) State whether you consider vibration to be indicative of 

structural weakness, giving reasons for your answer. 

(b] How would you attempt to reduce vibration when excessive ? 
(f) At the lowest number of vibrations possible, where would 

you expect to find the nodal points ? 

(d) What recent modifications in design are known to produce 
less vibration ? (50) 

71. Describe fully the method of conducting measured-mile 
trials and arriving at the measured-mile speed. 

State the possible sources of error to which such trials are 
liable, and how they are reduced. (40) 

72. How would you obtain the wetted surface of a ship of known 
form ? 

Quote any formula giving a close approximation to the wetted 
surface. 

What use can be made of the wetted surface when obtained ? 

The wetted surface of a ship of 6000 tons displacement being 
25,000 square feet, find the wetted surface of a vessel of similar 
form, but of 2000 tons displacement. (40) 

2 H 



466 Appendix. 

73. A vessel runs bow-on to a shelving beach ; investigate her 
stability, as compared with her condition when afloat. 

A box-shaped vessel, 100 feet long and 20 feet broad, floats at a 
draught of 6 feet forward and 10 feet aft, the metacentric height 
being T.\ feet. Find the virtual metacentric height when she just 
grounds all along on level blocks. (45) 

74. Sketch and describe the launching arrangements for a large 
ship, stating the dimensions of the vessel, declivity of the blocks 
and launching ways, and the pressure per square foot allowed on 
the surface of the ways. What is the meaning of the term 
" camber " as applied to the ground ways, and to what extent is it 
admissible ? 

The launching weight of a ship is 2800 tons, its centre of gravity 
is 9 feet abaft the midship section, and the fore end of the launching 
cradle is 120 feet before the midship section. When the midship 
section of the vessel is respectively o, 10, 20, 30, 40, and 50 feet 
abaft the after end of the ways, the corresponding buoyancy is 
respectively mo, 1310, 1530, 1770, 2030, and 2310 tons, and the 
distances of the corresponding centres of buoyancy abaft the after 
end of the ways are respectively 43, 51, 60, 68, 77^, and 86 feet. 

Construct to scale the corresponding launching diagrams, 
stating where the stern begins to lift, and the pressure on the fore 
poppet. Are the ways sufficiently long to prevent tipping ? (55) 

75. Investigate the value of the metacentric height of a vessel 
with free water in the hold. 

A mud hopper of box form is 200 feet long and 40 feet broad, 
the mud chamber being the amidships portion 50 feet long. When 
empty, the draught is 10 feet and the centre of gravity 15 feet 
above the keel. Find the metacentric heights when (a) Empty, 
() Discharge-port is open, and (c) Chamber is filled to a height of 
10 feet with sludge of specific gravity 2. (45) 

76. Describe in detail how you would proceed to fix the dimen- 
sions and underwater form of a combined passenger and cargo 
carrying steamer, having given the speed, length of voyage, 
maximum draught permissible, cargo capacity (both by measure- 
ment and dead- weight), number of passengers, and type of vessel. 
State what you consider satisfactory limits of stability for such 
a vessel as you select. (50) 



Appendix. 467 

1908. 
STAGE II. 

Instructions. 

You are permitted to answer only eight questions. 

You must attempt Nos. 32 and 33 ; also three questions in 
the Practical Shipbuilding Section, and one in the Laying Off 
Section. The two remaining questions may be selected from any 
part of the paper in this stage. 

PRACTICAL SHIPBUILDING. 

21. Describe, and show by sketches in section and side 
elevation, how an intercostal plate keelson (or vertical keel) is 
worked and secured in an ordinary transversely framed vessel 
with a flat-plate keel. (20) 

22. Show how a large transverse watertight bulkhead is built, 
stiffened, and riveted. 

How would you check the position of such a bulkhead at 
the ship, and how would you test its watertightness when 
completed ? (25) 

23. Sketch the portion of the midship section extending from 
the margin plate to the upper deck in a mercantile vessel, or, 
from the 4th longitudinal to the upper deck in a war vessel. Show 
the arrangement in detail, with scantlings of outer bottom plating, 
framing, stringers, etc., and name the type of vessel selected. 

(22) 

24. Sketch and describe the construction of the stem of any 
large vessel, showing in detail its connections to the keel, decks, 
and shell plating. Name the type of vessel selected, and state 
the material of which the stem is made. 

What tests would you apply to ascertain the fitness of the 
stem ? (20) 

25. Sketch a good shift of butts for the tank top or inner 
bottom plating of a vessel, with reference to the butts of the 
girders or longitudinals. Give details of the butt and edge con- 
nections of the plating, and state the spacing of the frames, length 
of plates, etc. (22) 

26. Sketch in detail, and describe a good arrangement of 
boiler-bearers, giving scantlings, etc., and name the type of vessel 
selected. (20) 



468 Appendix. 

27. Compare the qualities of " mild " and " high-tensile " steel, 
and state the tests you would expect high-tensile steel plates to 
satisfactorily comply with. 

In what parts of the structure of a ship would it be of advantage 
to fit high-tensile steel, and what precautions should be taken in 
working it ? (23) 

28. What permanent arrangements are made for protecting the 
side of a ship which has frequently to lie alongside quay walls, etc. ; 
such as cargo vessels, tugs, etc. 

Sketch and describe the construction of any suitable arrange- 
ment, giving scantlings and particulars of the fastenings, and state 
where such an arrangement should be fitted. (20) 

LAYING OFF. 

29. Define the following terms, viz. " Length over all," 
" Length between perpendiculars," " Moulded breadth," " Moulded 
depth," " Rise of floor," " Dead flat," " Tumble home," " Depth of 
hold," " Camber of beam," and " Bilge diagonal." (20) 

30. How would you make a beam mould ? Show how to obtain 
the beam end line in the sheer and body plans, taking into account 
the round-up and sheer of the deck. (22) 

31. What information would be required, and how would you 
proceed to demand the materials, for rapidly building a steel 
vessel, indicating the order in which you consider the materials 
should be demanded ? 

What margins are allowed over the actual dimensions 
required? (25) 

DRAWING. 

32. The given sketch represents not to scale five equidistant 
sections, obtained by taking measurements from the outside of 
a vessel in dry dock. Offsets to the various water lines, level 
lines, and bow lines are given in the table accompanying the 
drawing. 

With the given offsets and particulars on the drawing, draw 
the five sections neatly, in pencil, to a scale of \" equals i foot. 

(35) 
CALCULATIONS. 

33. Write down and briefly explain the rules in common use 
in ship calculations, for finding the areas of plane surfaces and 
volumes of displacement. 

The semi-ordinates of the boundary of a deck of a vessel 



Appendix. 469 

are :~o*5, 4*5, 88, io'o, 8'2, 3-8, and 0-4 feet respectively, including 
the end ordinates. The length of the vessel being 85', find the 
area of the deck, and the position of its centre of gravity. (25) 

34. A deck of a vessel is composed of flush plating &" thick, 
secured to channel bar beams 8" x 4" x ", spaced 3' 6" apart. 

Calculate the weight of a part of the deck 63' long by 10' wide, 
including rivets, but omitting edge strips and butt straps. (20) 

35. What are the curves of " displacement " and " tons per inch 
immersion," and what are their uses ? 

The area of a ship's load water plane is 6050 square feet, the 
body below is divided by equidistant horizontal sections 3' apart, 
whose areas are 5500 ; 4750 ; 3500 ; 2050 ; 1000 ; and 250 square 
feet respectively. 

Find the tons per inch at each water plane, and plot the curve 
of tons per inch, on the squared paper supplied. 

What is the total displacement of the vessel ? (23) 

36. Explain, in detail, a method of determining graphically the 
displacement of a vessel of given form. (22) 

STAGE III. 

Instructions. 

You are permitted to answer only eight questions. 

You must attempt No. 52. The remaining questions may be 
selected from any part of the paper in this stage, provided that one 
or more be taken from each section, viz. Practical Shipbuilding, Lay- 
ing Off, and Calculations ; but you must not attempt more than 
three questions ', including No. ^from the Calculations section. 



PRACTICAL SHIPBUILDING. 

41^ Describe with the aid of rough sketches, the fittings neces- 
sary for efficiently working the anchors of any vessel with which you 
are acquainted, naming the type of vessel selected. (35) 

42. Describe the difference between an " ordinary " and a 
"balanced" rudder, and tate why the latter are sometimes fitted 
in ships. 

Sketch a sternpost suitable for a balanced rudder, and show how 
it is connected to the keel, decks, and shell-plating. (35) 

43. Roughly sketch the midship section of any vessel with which 
you are acquainted, naming the type of vessel selected. 

Indicate the scantlings of the various parts comprising the 
section. (35) 



4/0 Appendix. 

44. Sketch and describe in detail the construction of an accom- 
modation or gangway ladder, showing how it is raised and stowed. 

(35) 

45. What precautions are taken in oil-carrying steamers to avoid 
risk Of explosion ? 

How are the bulkheads of such vessels constructed? Give 
scantlings of plates, etc., and size and spacing of rivets. (35) 

46. Describe, with sketches, the construction of a horizontal 
punching machine, and explain how the machine can be adapted 
to do riveting, beam-bending, and angle bar cutting. (35) 

47. Explain how the Rules of Lloyd's Register determine the 
scantlings of a three-decked merchant steamer, distinguishing 
between the transverse and the longitudinal portions of the 
structure. 

In a vessel as above, it is desired for convenience of stowage 
to omit the " hold " beams. Describe the modifications you would 
adopt to the ordinary construction consequent upon that omission. 

(40) 

48. Describe, with sketches, the method of coaling a large ship, 
such as an Atlantic liner or a warship, from a collier or barge 
alongside. 

Show, by a sectional sketch of the vessel, how the coal is passed 
from the upper or coaling deck, to the bunkers. (35) 

49. Sketch and describe the construction of a steel deck-house. 
In the case of a vessel which has a long bridge-house to be 

worked amidships, how should this bridge-house be constructed in 
order that it maybe made an efficient pa/t of the structural strength 
of the vessel? (35) 

LAYING OFF. 

50. Distinguish between a " ribband," a " harpin," and a " sheei 
harpin." 

Show how to lay off and obtain the bevellings of a sheer 
harpin. (33) 

51. The lines of a steel vessel, sheathed with wood, having been 
given to the outside of sheathing, show how you would obtain Ithe 
body plan to outside of framing (i) approximately, and (2) 
accurately. (35) 

CALCULATIONS. 

52. The given sketch represents part of the fore body of a ship. 
Calculate the displacement in tons, and the vertical position of the 
centre of buoyancy of the form represented by the sketch, between 



Appendix. 47 j 

the water-lines A and B, 12' apart, and between the sections C and 
D. The sketch given is on a scale of \" equals I foot, and the 
sections are spaced 20' apart. Four water-lines at depths of 3', 6', 
9', and 10' 6" below A water-line are to be introduced between A 
and B for the purpose of the calculations. 

Ordinates to be measured to the nearest first decimal place. 

(45) 

53. State and prove Simpson's ist Rule, for approximating to 
the area and centre of gravity of a plane surface. 

The equidistant half-ordinates of a water-plane being 3*0, 5*4, 
7- 1, 9-32, 12-2, 14-17, and 19-5 feet respectively, and the length of 
the base being 84*0 feet, find the area of the water-plane, and the 
transverse position of the centre of gravity of half the water-plane. 

(40) 

54. The half-ordinates of a portion of a ship's deck, covered with 
&" plating, are 4-2, 9-36, 12-3, 14-84, 16-5, 17-53, and 187 feet in 
length respectively, the common interval being 1 5 feet. 

Calculate the weight of the beams, plating, planking, and 
fastenings, etc., for this part of the deck, the beams being 8" X 4" 
x j$y" channel bar, spaced 3' 6" apart, and the plank being of pitch 
pine 3$" thick. 

Estimate the cost of laying the deck with planks 6" wide at 
1\d. per foot run. (35) 

55. What is the ultimate shearing and tensile stresses of mild 
steel rivets and plates respectively ? 

The shell plating of a vessel is formed of plates 50" wide and 
^" thick, worked on the raised and sunken system ; the spacing 
of the rivets in the frames are 7 diameters apart, and in the 
boundary angles of the watertight bulkheads 4 diameters apart. 
Sketch an arrangement you would make in order that the strength 
of the shell plating in wake of the bulkheads and ordinary frames 
shall be approximately the same. Show, by calculations, that 
your arrangement is a good one. (35) 

56. Define " centre of gravity," " centre of buoyancy," " centre 
of flotation," " metacentre." 

A vessel 140' long, and whose body plan half-sections are 
squares, floats with its sides upright, and the centres of all the 
sections lie in the plane of flotation. The lengths of the sides of 
the sections, including the end ordinates, are o'8, 3 '6, 7'o, 8'o, 6*4, 
3'o and 07 feet respectively, equispaced. 

Calculate the distance between the centre of buoyancy and the 
transverse metacentre. (35) 



472 Appendix. 

HONOURS. 

Instructions. 

You are not permitted to answer more than eight questions. 

Note. No candidate will be credited with a success in this 
examination who has not obtained a previous success in Stage III. 
or in Honours, of the same subject. 

Those candidates who do well in the following paper will be 
admitted to a practical examination held at South Kensington or 
some other centre. Candidates admissible to that examination will 
be so informed in due course. No candidate will be classed in 
Honours who is not successful in the Practical Examination. 

61. Having given the value of six equidistant ordinates of a plane 
curve, deduce a formula that will give the area of the surface lying 
between the extreme ordinates and the curve. 

Four consecutive polar radii of a curve, taken in order, are 
io'9, 1 1 '6, 13x3, and 14/1 feet; the common angular interval 
between them is 15 degrees. Find the area, in square feet, included 
between the curve and the extreme polar radii, and prove the rule 
you use. (45) 

62. Obtain an expression giving the height of the longitudinal 
metacentre above the centre of buoyancy. What use is the 
information when obtained for any particular vessel ? 

Draw, to scale, the ordinary metacentric diagram for a vessel 
whose uniform section throughout her length is a quadrilateral of 
breadth 50' at the load line and 25' at the keel, the draught of 
water being 20'. (45) 

63. Under what circumstances may it be expected that the 
cargoes of vessels will shift ? 

In a cargo-carrying vessel, the position of whose centre of 
gravity is known, show how the new position of the centre of 
gravity, due to a portion of the cargo shifting, may be found. 

A ship of 4800 tons displacement, when fully laden with coals, 
has a metacentric height of 2*6 feet. Suppose 120 tons of coal to 
be shifted so that its centre of gravity moves 19 feet transversely 
and 5 feet vertically, what would be the angle of heel of the vessel, 
if she were upright before the coal shifted ? (45) 

64. Prove that for any floating body revolving about an axis 
fixed in direction, positions of maximum and minimum stability 
occur alternately. 

Investigate all the positions of equilibrium for a square prism 



Appendix. 473 

of uniform density revolving about a horizontal axis, assuming its 
density to be three-fourths that of the fluid it is floating in. (50) 

65. Quote Moseley's formula for the dynamical stability of a 
floating body, and prove that the value of the dynamical stability 
obtained from that formula is identical with that obtained by 
integrating the curve of statical stability. 

A vessel of constant rectangular section is 260' long, 30' broad, 
30' deep, and draught of water 15'. The metacentric height of the 
vessel being 2-5', find (i) the statical stability, 0</(2)the dynamical 
stability of the vessel when she is inclined at 45 degrees. (50) 

66. A box-shaped vessel 420' long, 72' broad, and draught of 
water 24', has a compartment amidships 60' long, with a water- 
tight middle line bulkhead extending the whole depth of the vessel. 
Determine the angle of heel caused by the ship being bilged on 
one side abreast this bulkhead, the centre of gravity of the vessel 
being 23' above the keel. 

To what height should the transverse bulkheads at the ends of 
the bilged compartment be carried, so as to confine the water to 
this part of the vessel ? 

If the bilging be caused by a collision, making a hole 1*5 square 
feet in area at a depth of 18' below the load water-line of the 
vessel, in wake of the compartment referred to above, and the 
pumps be in working order, calculate the capacity of the pumps 
required to just keep the leak under. (45) 

67. Define " freeboard," and state what determines it. Describe 
the arrangement of the tables giving " freeboard." 

How is the statutory deck-line marked ? 

Distinguish between "flush-deck," " spar-deck," and " awning- 
deck" vessels. How is the freeboard determined in each case ? (50) 

68. Prove the relation which exists between the load curve and 
the curves of shearing force and bending moment. 

A vessel 300' long has a uniform section below water. The 
weights of hull, machinery, and cargo are 840, 300, and 300 tons 
respectively. The weight of machinery extends uniformly over 
rd of the length amidships, and the weight of cargo extends 
uniformly over th of the length from each end. The weight of 
hull curve is of the form 




I 

1H 



I 



->k 




FIG. 157 



t 



474 Appendix. 

Draw the curves of shearing force and bending moment and 
state their maximum values when the vessel is at rest in still 
water. ( 45 ) 

69. The effective part of the transverse section of a vessel amid- 

I ships is represented by the dia- 

gram (Fig. 1 58), the vessel being 
<o I 42' broad and 28' deep. 

Find the maximum tensile 
and compressive stresses when 
the vessel is subjected to a sag- 
ging bending moment of 60,000 
foot-tons. The plating shown in 
the diagram to be taken as i" 

u 14.' >{ 14.' I J< |4/ J thick, and no allowance need be 

' made for rivet holes or laps of 
FlG - '5 8 ' plating. 

Assuming I* and I v are the moments of inertia of a section 
about axes at right angles to each other, deduce a formula for 
finding the stress on the section at any point when a vessel is 
inclined at an angle to one of the axes. (50) 

70. Enumerate the component parts of the total resistance to 
propulsion of a ship. What is the relative importance of these 
component parts at (i) low speeds, and (2) at high speeds? 

A ship, 290' long, 45' beam, 17' 6" draught, and 3200 tons 
displacement, steams 17^ knots. Find the horse-power necessary 
to overcome frictional resistance, having given that the resistance 
varies as the 1-83 power of the velocity, and that in fresh water, at 
a speed of 10 feet per second, the average resistance for a length 
of 50' is 0*246 Ibs. per square foot, whilst over the last square foot 
the resistance is 0*232 Ibs. (45) 

71. Explain in detail how the indicated horse-power for a new 
ship is estimated. 

A model of a vessel, 400' x 65' x 24' draught, of 8560 tons 
displacement, is run, and the curve of E.H.P. on a base of speed 
of ship is 3250, 4035, 5020, 6195, and 7660 E.H.P. for 16, 
17, 18, 19, and 20 knots respectively. Make an estimate of the 
I.H.P. of a ship of 16,000 tons, of similar form, for speeds of 20 and 
21 knots, and give the dimensions of the new ship. (45) 

72. The draught of water, the desired speed, and the load to be 
carried being given for a new design, state in detail how you would 
obtain the approximate dimensions of the ship. 

Obtain suitable dimensions for a vessel to carry uoo tons of 
cargo on a limiting draught of 21', the speed of the vessel to be 



Appendix. 475 

12 knots, with coal sufficient for a voyage of 1500 miles, and 
300 tons of passengers and stores. (50) 

73. Deduce a formula for the period of a ship whose rolling is 
unresisted and isochronous. 

A vessel of 13,500 tons displacement has a metacentric height 
of 3*5 feet and a period of 8*5 seconds. Find the period of rolling 
when 600 tons of coal are added each side of the vessel in a bunker 
21' deep and 9' wide, the centre of gravity of the bunkers being 
n' below the original centre of gravity of the ship and 26' out 
from the middle line. The vessel has a horizontal curve of 
metacentres over the limits of draught corresponding to the above 
conditions. (50) 

74. Define the terms "effective wave slope" and "virtual 
upright." Explain under what circumstances the rolling of a 
vessel amongst waves is likely to be most severe, and state what 
resistances are in operation to prevent overturning in such critical 
cases. 

What conditions should be fulfilled in a ship to make her easy 
in her rolling at sea ? (45) 

75. Discuss the distinctive features of torpedo vessel design. 
What are the most recent developments in the design of this class 
of vessel in this country ? 

What is the effect of depth of water upon the speed of a 
vessel ? 

State the deductions that have been made from recent trials 
with vessels in shallow water. (50) 

76. Describe how to construct a set of lines, having given the 
type of vessel, dimensions, displacement, and the position of the 
longitudinal centre of buoyancy. 

Having obtained the sheer drawing of a vessel, how would you 
proceed to obtain the structural midship section on the under- 
standing that the vessel is to be built to meet Lloyd's require- 
ments ? (50) 



476 Appendix. 



1911. 
STAGE II. 

Instructions. 

You are permitted to answer only eight questions. 

You must attempt Nos. 32 and 33 ; also three questions in the 
Practical Shipbuilding Section, and one in the Laying Off Section. 
The two remaining questions may be selected from any part of the 
paper in this stage. 

PRACTICAL SHIPBUILDING. 

21. Describe briefly, with sketches, the tools and appliances 
used in ordinary shipyard work. State the advantages and dis- 
advantages of machine-riveted work, as compared with hand- 
riveting. (20) 

22. Sketch, and describe, the construction of a transom frame, 
showing how it is connected to the other parts of the vessel. Show 
how the frames of the stern or counter are connected to it. (20) 

23. For what purposes are web-frames fitted in ships ? 

Sketch, and describe, any arrangement of web-frames asso- 
ciated with side stringers. A plan, section, and elevation to be 
shown representing the arrangement described. (22) 

24. How would you construct a small hatchway in an upper 
deck which is planked, but not plated ? (20) 

25. Show by sketches, and describe, an efficient arrangement 
of butts of an upper deck stringer plate, stating spacing of beams, 
and give details of the butt connections. State the scantlings of 
the plating, the size and pitch of rivets in edges and boundary 
bar. Also, show the relative position of the butts of the adjacent 
sheer strake, and boundary angle. (23) 

26. For what purposes are bilge keels fitted to ships ? 
Describe in detail, with sketches, the construction of a bilge 

keel, showing how the several parts are connected to each other, 
and to the bottom plating. State the scantlings of keel, diameter 
and pitch of rivets, etc. How is the bilge keel lined off at the 
ship ? (25) 

27. A steel plate is intended to be fitted to the side of a ship, 
where great curvature and twist exists ; describe the whole of the 
operations in connection with the plate, from the time it enters 
the yard until it is finally riveted in place. (22) 



Appendix. 477 

28. Describe, with sketches, how a ship is supported during 
building, and roughly mark the position of the ribbands in relation 
to the lands of bottom and deck plating. How is a ship's form 
checked during building? Describe how frames, and lands of 
shell plating, are faired at the ship. (20) 

LAYING OFF. 

29. What is meant by " fairing " the body plan ? How is a 
line tested for fairness ? Describe, with sketches, the appearance 
of the following lines, in the profile, half-breadth and body plans, 
viz. : " transverse frames,'' " diagonals," " bow and buttocks," 
" levels " or " water-lines," and " deck " lines. 

Which of the above lines represent the true form of the ship, 
in the particular plans ? (25) 

30. Show how to find the point where a straight line, not 
parallel to any of the planes of projection, viz. : sheer, half 
breadth, or body plan, would, if produced far enough, cut the 
surface of a ship of known form. (22) 

31. Show, by rough sketches, the general appearance of the 
sight edges of the outer bottom plating in the fore and after body 
plans of a vessel. A sufficient number of frames in each body 
should be shown, so as to indicate thereon the character of the 
sight edges. 

Show also, on the sketch, the trace of a keelson, inner bottom 
frame line, and a girder or longitudinal. (20) 

DRAWING. 

32. The given sketch represents (not to scale] the stem and 
part of the framing, etc., of the fore part of a vessel. Draw it 
neatly, in pencil, to a scale of " equals I foot. (35) 

CALCULATIONS. 

33. The area of a ship's loadwater-line section is 13,200 square 
feet, and the areas of other parallel sections 3' apart, are as 
follows, viz. : 12,700, 12,000, 11,100, 10,000, 8,200 and 6,000 square 
feet respectively. Neglecting the volume below the lowest section, 
calculate (i) the tons per inch immersion at each water-plane, and 
(ii) the total displacement of the vessel. 

Construct, on the squared paper supplied, the curve of tons 
per inch. (25) 



478 Appendix. 

34. The half-ordinates of the transverse section of a coal bunker 
of uniform section are as follows, viz. : 31*0, 31*3, 30*8, 29^0, and 
24-6 feet respectively, the ordinates being spaced 5' apart. The 
length of the bunker is 25 feet. 

On the basis of the coal being stowed only up to the level of 
the underside of beams, which are 5" deep and spaced 4' apart, cal- 
culate the weight of coal that can be so carried in the bunker. (20) 

35. What is meant by the shear of a rivet ? Explain clearly, 
with sketches, the difference between "single" and "double" 
shear. 

What is the single shear strength of a f " diameter mild steel 
rivet? 

Two test bars, of circular section, \" diameter, are prepared 
from the following materials, viz. : (a) mild steel, and () rolled 
Naval brass, or yellow metal : what breaking force would you 
expect the testing machine to register when the bars are broken ? 

What elongation would you expect, in each case, on a length 

Of 2"? (2 3 ) 

36. Explain why vessels passing from salt water to fresh water 
change their draught. What condition must be fulfilled in order 
that a vessel may not change trim in going from fresh to salt 
water, or vice versa f 

A box-shaped vessel is 175' long, 30' broad, 20' deep, and floats 
at a uniform draught of 8' in salt water. Calculate the mean 
draught when the vessel is floating freely in fresh water. (20) 



STAGE III. 

Instructions. 

You are permitted to answer only eight questions. 

You must attempt No. 52. The remaining questions may be 
selected from any part of the paper in this stage, provided that one 
or more be taken from each section, viz., Practical Shipbuilding, 
Laying Off, and Calculations ; but you must not attempt more than 
three questions, including No. ^from the Calculations section. 

PRACTICAL SHIPBUILDING. 

41. The drawings of a large vessel having been received by 
the Builders, describe the preliminary work necessary in order to 
ascertain whether the ship can be built and launched from the 
building slip intended to be used. 



Appendix. 479 

A slip is 480' long from the sill to the foremost block, and has 
a declivity of f" to i foot. The building blocks are to be 
laid at a declivity of f" to i foot, and the launch is to be at a 
declivity of " to i foot. Determine the height of the foremost 
block. (35) 

42. Enumerate the principal transverse stresses experienced 
by ships. Describe how they are set up, and the provision made 
to meet them. 

In the case of a large machinery, or cargo, hatch in the deck 
of a vessel, show what means are adopted to compensate for the 
loss of transverse strength due to cutting the deck, etc. (40) 

43. Show how a transverse watertight bulkhead extending 
the whole depth of the vessel, in a merchant ship of say 50' beam 
is built, stiffened, and riveted. State the scantlings, and size and 
pitch of rivets. How would you test the watertightness of such a 
bulkhead ? 

State the number, and positions, of the watertight bulkheads 
required by Lloyd's Rules for a steamship 400' long. (35) 

44. Roughly sketch, giving figured dimensions of, an anchor 
davit or anchor crane to be used in connection with the anchor 
arrangements of a vessel, and show all the necessary fittings, etc., 
for working the crane. State what tests are sometimes applied to 
such davits or cranes. (35) 

45. Describe, with rough sketches showing transverse and 
longitudinal midship sections, the method of construction of a 
large wood pulling boat forming part of the equipment of a ship. 
Indicate the materials and the fastenings used. (35) 

46. Having given the particulars and materials for building an 
ordinary steel lower mast of a ship, how would you proceed with 
the construction so as to ensure its being made to its correct 
form ? Show a section of the mast, and the disposition of the 
butts of the plates. 

State the sizes of the plates, rivets, etc., and show the riveting 
at a butt. (35) 

47. Sketch and describe the construction of a boat's davit, with 
fittings complete, and state the object of the various fittings shown. 
How is the position of the davits fixed in relation to the boat ? 

Show the method of securing the boat in-board, and of rapidly 
getting the boat in the water. (35) 

48. Describe in detail, with sketches, the special arrangements 
for towing purposes, as fitted in a large tug boat. 

State where such fittings are placed, and why ? (35) 

49. Sketch, in detail, the arrangements of a Seaman's head, or 



480 Appendix. 

water-closets, for a ship's crew of say 150, or more. Show how 
the soil pipes are arranged. 

What ventilation arrangements are made for such spaces, in 
the case which you select ? Name the type of vessel selected. 



LAYING OFF. 

50. Show how to lay off the stern, and obtain the true expanded 
form of the stern plating above the knuckle line and abaft the 
transom, in the case of a vessel having such part formed by the 
rolling of a cylinder about the knuckle line. (35) 

51. Describe the method adopted for laying off a longitudinal, 
or tank margin plate, in cases where (i) there is little twist, and 
(ii) where considerable twist occurs. 

Describe, in detail, the information supplied to the workmen in 
both these cases. (35) 

CALCULATIONS. 

52. The given sketch represents part of the after-body of a 
ship. Calculate the displacement in tons, and the vertical position 
of the centre of buoyancy, of the form represented by the sketch, 
between the waterlines A and B, spaced 10' 6" apart, and between 
the sections C and D which are 60' apart. The sketch given is 
to a scale of " equals I foot. 

Three waterlines, at depths of 3' 6", 7' o", and 8' 9'* below A 
waterline, are to be introduced between A and .Z?, for the purpose 
of the calculations. 

Ordinates are to be measured to the nearest decimal place. 

(45) 

53. Define the terms : " centre of flotation," " centre of 
buoyancy," and " metacentre." 

A prismatic log of wood, of specific gravity 075, whose uniform 
transverse section is that of an isosceles triangle, floats in water 
with the base of the section horizontal and vertex upwards. Find 
the maximum vertical angle of the section for these conditions to 
hold. (35) 

54. Describe fully the method of making and arranging the 
various calculations on a displacement sheet, and state fully what 
information is usually shown thereon. 

Explain the relation which exists between a curve of tons per 
inch and the corresponding curve of displacement, and show how 
either curve may be derived from the other. 



Appendix. 481 

Distinguish between displacement and deadweight scales, and 
show clearly how each is generally arranged. (40) 

55. What conditions must be fulfilled in order that a vessel 
may not change trim in going from fresh to salt water, or vice 
versa ? 

A rectangular vessel, 300' long and 40' broad, floats at a 
draught of 10' forward and 12' aft in sea water. Find the draught 
at which she will float in fresh water weighing 62^ Ibs. per cubic 
foot, the centre of gravity being situated in the original waterline. 

(35) 

56. Define the term " Statical Stability." Show, by means of 
a diagram, the forces acting on a ship when inclined. What is 
the " righting lever " ? 

Sketch a typical statical stability curve, indicating the principal 
points of importance on it. (35) 



HONOURS. 
Instructions. 

You are not permitted to answer more than eight questions. 

Note. No candidate will be credited with a success in this 
examination who has not obtained a previous success in Stage 3, 
or in Honours, of the same subject. 

Those candidates who do well in the following paper will be 
admitted to a practical examination held at South Kensington or 
some other centre. Candidates admissible to that examination 
will be so informed in due course. No candidate will be classed 
in Honours who is not successful in the practical examination. 

61. Deduce a rule for finding the area of a curvilinear figure? 
by means of 5 ordinates, so spaced that the area of the figure is a 
multiple of the sum of the ordinates. 

Five consecutive polar ordinates of a curve, taken in order, are 
5'o, 5 '2, 57, 6'4 and 7*3 feet respectively, and they are spaced at a 
common angular interval of 5 degrees. What is the area, in square 
feet, included between the curve and the extreme polar radii ? 

(So) 

62. State fully, and prove, the conditions of equilibrium of a 
floating body. 

Define the terms " stable," " unstable," " neutral," and " mixed " 
equilibrium. 

Show that, in the case of a floating body, the equilibrium is 

2 r 



482 Appendix. 

stable when the distance between the centre of gravity and centre 
of buoyancy is a minimum, and unstable when that distance is a 
maximum. Discuss the relation between the number of positions 
of stable and unstable equilibrium. (50) 

63. A vessel is inclined about an axis, in the water-line plane, 
which makes an angle, other than a right angle, with the longi- 
tudinal middle line plane of the ship. Obtain an expression con- 
necting the metacentric height under these conditions with the 
transverse and longitudinal metacentric heights of the ship. 

A box-shaped vessel is 80' long, 20' wide, and floats at a 
draught of water of 10'. Find the value of the distance between 
the centre of buoyancy and the metacentre, for inclinations about 
an axis coincident with a diagonal of the rectangular waterplane. 

(45) 

64. Describe, in detail, how an inclining experiment is carried 
out. What observations are made? Show how to deduce. the 
correct height of the centre of gravity, if loose water was lying in 
the bilges when the observations were made* How would you 
determine the amount of ballast to be used on an inclining 
experiment ? 

What special calculations would you make, if the vessel at the 
time of inclining were considerably out of her normal trim? (45) 

65. Investigate, and sketch, the metacentric diagram for a 
vessel of constant parabolic section throughout, and show that in 
such a vessel the presence of free water in the hold, in any number 
of compartments, leads to an increase of stiffness. 

Draw, roughly, the metacentric diagrams for three distinct 
types of modern vessels, naming the types chosen. Figure on the 
diagrams the values of the metacentric heights for the load and 
light conditions in each case. (45) 

66. State, and prove, Atwood's formula for the moment of 
statical stability of a floating body when inclined at any angle from 
the upright. State clearly how to determine the sign of the 
moment of the correcting layer. 

A prismatic vessel, 100' long, has a transverse section formed 
of a rectangle, height 10' and breadth 20', resting on the top of a 
semicircle of radius 10'. The centre of gravity is 3' above the 
keel, and the draught of water is 10'. Find the volume of the 
correcting layer, and the righting moment when the vessel is 
inclined 45, the displacement being unaltered. (45) 

67. Define " Reserve Dynamical Stability/' and explain its 
importance in the case of sailing ships. 

Define " power to carry sail," as applied to sailing ships, and 



Appendix. 483 

explain clearly why it is usually less in small ships than in 
large, and why in yachts a small value can generally be safely 
accepted. (50) 

68. Define " Freeboard," and " Range of Stability," and state 
what determines each. 

Explain clearly why, in general, high freeboard is conducive 
to a long range of stability, and low freeboard to a short range. 
Show, by simple illustrations, that in certain cases a low freeboard 
may be associated with a considerable range, and a high freeboard 
with a short range. (45) 

69. Find an expression for the heel produced in a vessel by 
flooding a compartment extending to the upper deck, and bounded 
by two transverse bulkheads and a middle line bulkhead. 

A vessel of square transverse section, 40' broad and deep is 
270' long and floats at a uniform draught of 20'. It has 8 equi- 
distant watertight transverse bulkheads, excluding the ends, and a 
longitudinal middle-line bulkhead over the midship portion. Find 
the heel produced by bilging the centre compartment, on one side 
of the middle line, the original metacentric height of the vessel 
being 5'. 

70. Prove the relation which exists between the curves of loads 
and shearing forces. 

Plot a shearing stress curve for a rectangular beam 12" deep 
and 8" wide, at a section where there is a shearing force of 180 
tons. What is the maximum shearing stress at the section ? (50) 

71. State the assumptions upon which the trochoidal wave- 
theory is based, and the propositions and conditions which must 
be fulfilled. 

How would you construct a trochoidal wave-profile of given 
dimensions ? 

Show clearly how to obtain the supporting force per foot, taking 
into account wave-pressures. What is the effect upon the maximum 
stresses caused by taking wave pressures into account, and why ? 

(50) 

72. Deduce the equation of motion for a vessel rolling unre- 
sistedly in still water. Obtain its solution, making the necessary 
assumptions. Show that the motion is oscillatory, and deduce a 
formula giving the period of oscillation of a vessel. (50) 

73. A vessel has 12 guns capable of firing on each broadside, 
the mean height of the centre of guns being 26' above the water- 
line, and the draught of water 27'. The ship has a displacement 
of 22,500 tons, and a metacentric height of 5'. Taking the weight 
of the shot as 850 Ibs., powder 270 Ibs., and muzzle velocity of 



484 Appendix. 

projectile as 2,900 feet per sec., estimate the maximum angle of 
roll of the ship caused by the simultaneous firing of all the guns 
on the broadside, omitting any resistance to the heel. The period 
of oscillation of the ship in still water is 9 seconds. (50) 

74. Sketch six different types of merchant steamships, naming 
the several decks and part decks in each case, as well as the name 
of the type. Explain the particular advantage of each type, and 
trace the evolution of the modern merchant steamer from the 
original flush one-deck type. 

75. What is the " Admiralty displacement coefficient of speed " ? 
State the assumptions on which it is based. 

How is it obtained for any particular vessel, and what use is 
made of it ? What is its value in three distinct types of vessels ? 
Name the types selected. 

Show that this coefficient is the same for two similar ships at 
"corresponding" speeds, supposing that the engines, etc., work 
with the same efficiency. 

What is the value of the coefficient for sea work, as compared 
with that deduced from trial trips ? (50) 

76. What are the most important developments, from a 
designer's point of view, that have taken place in recent years, in 
any two of the following types of vessels, viz. : 

(a) Ships of the Mercantile Marine ; 

(b) Motor Boats of high speed ; 

(c) Armoured Ships of War ; 

(d) Torpedo Boat Destroyers ? (50) 



Appendix. 



485 



ANSWERS TO QUESTIONS. 



EXAMINATION 

No. 

12. 449 tons. 

13. 674 square feet. 

14. 420 ; see p. 28. 

16. See p. 37 ; 240 Ibs. ; 3 inches. 

29. 5,459 cubic feet ; 2-29 feet 

below No. i W.L. ; 49*1 
feet from fine end. 

30. 8280*8 cubic feet j 49-4 feet 

from fine end. 

31. 5-48 feet. 

32. 2| tons if of steel. 

46. 797 tons; 1515 tons. 

47. 4-38 feet from lo-feet W.L. ; 

1 1 6-6 feet from No. i 
section. 

48. 1 3*49 feet. 

49. 297 tons ; 346 tons (assuming 

28 tons per square inch). 

50. 8 inches forward, 7 inches aft. 
61. About 50 tons if of steel ; 0-2 

foot forward of midships. 
53. Take two consecutive sections 
of beam K and K', distance 
A* apart ; w = load per 
foot run ; F and F + AF 
are shearing forces at K 
and K' respectively ; M 
and M + AM are 'bending 
moments at K and K 7 
respectively. 
Consider the equilibrium of 

beam between K and K'. 
Vertical forces up, F + AF ; 
,, ,, down, F and 

iu X A* ; 

.% F + AF = F + (w X A#) 
or AF WY. A* 
and F = ^wdx in the 
limit. 



PAPER, 1902. 

No. 

Also for equilibrium 

M + AM = M + (F X A*) 
or AM = F x A# 
and M = JYdx in the 

limit. 

54. The equations to the curves 
of weight and buoyancy re- 
ferred to the base-line and 
one end are as follows : 
Weight 

, = 6- (/.*-**) 

Buoyancy 



58. 



A being the area of each, 

and / the length ; 

from which the curves of 

shearing force and bending 

moments may be obtained 

by a process of successive 

integration. 

Maximum shearing force at 
about T 3 S length from either 
end = T 5 weight about. 

Maximum bending moment 
amidships = ^ (weight x 
length} about. 

Take consecutive normals 
to the locus of centres of 
buoyancy at and 6 + A0, 
BR' being perpendicular to 
the latter normal from B, 
cutting B^ in R". Then 
RR" is the increment of 
B t R, i.e. A^R) and RR" 
also equals (BR X A0) ; so 
that 
A(B t R) = BR x A0 



486 



Appendix. 



No. 



59. 
61. 



Proceeding to the limit 

</(B t R) = BR x dO 
and therefore by integrating 



B V R = J BR . dQ 

A curve of GZ's enables a 
curve of BR's to be plotted, 
and the area of this curve 
up to a given angle (angles 
in circular measure) will 
give B,R, and so enable 
the position of the centre of 
buoyancy at that angle to 
be obtained. 

52 tons about. 

On account of the orbital 



No. 



03. 
65. 



motion of the particles of 
water in a wave, the virtual 
buoyancy is less in the crest 
portion and greater in the 
trough portion than at the 
same depth below the sur- 
face in still water. Calcu- 
lations, taking this into 
account, show that the 
bending moment is less 
than when calculating as 
described in the text. 

See a paper by the late Mr. 
T. C. Read, Jnst. Nov. 
Arch, for 1890. 

ii} tons per square inch about. 

360 E.H.P. nearly. 



EXAMINATION PAPER, 1905. 



No. 
33. 

35. 

53. 

54. 



55. 



5461 tons, 8 '04 feet below 

L.W.L. 
7860 square feet, 167 feet from 

first ordinate. 
3200 tons. 
About twenty $-inch rivets 

disposed in lozenge-shaped 

lap. 
2-13 feet. 



No. 
61. 
62. 
68. 



69. 

72. 
73. 



2-67 feet. 

2-3 feet. 

Ordinates of stability curve 1*7 
sin 6, range 180, 270 foot- 
tons. 

7800 I.H.P., assuming I.H.P. 
oc V 4 . 

12,000 square feet. 

07 foot. 



EXAMINATION PAPER, 1908. 



No. 
52. 



61. 
63. 
66. 
68. 



This question had to be done 
in two parts : (i) upper 9 
feet by Simpson's second 
rule, (2) lower 3 feet by 
Simpson's first rule, and 
the portions combined. 

60 '4 square feet. 

10 degrees. 

See example 24, Chapter V. 

The following are the curves 
required: S.F.max = 60 



No. 

tons, B.M.max = 1740 foot 
tons. 

69. See example 16, Chapter VII. 

70. 1420 about. 

(50 x 0-246) + (240 X 0-232) 
J ~ 290 

X i '025 = 0-241 at 10 f.s. 

71. 10,400, 12, 700,492' x 80' x 29-5'. 
73. See example at end of Chapter 
IX. 



Appendix. 



487 




FIG. 



EXAMINATION PAPER, 1911. 



No. 

53. 



If D is depth of section and 
the semi-vertical angle, 
then 



B above base = \ . D 

M above base = | . D + J . D . tan'0 

G above base = \ . D 

Equating the latter two ex- 
pressions tan 2 = I, from 
which 6 is 45 and the vertical 
angle should not be less 
than 90. 
55. A similar example to that 



No. 



61. 



63. 
66. 

69. 
71. 
73. 



worked out at the end of 

Chapter IV. 
TchebychetPs rule worked 

out similarly to that for 4 

ordinates in Appendix A. 
See end of Chapter V. 
This is worked out fully, 

Example 27, Chapter V. 
Similar to Examples 24, 25, 

26, Chapter V. 
See " Smith " correction, 

Chapter VII. 
See end of Chapter V. 



488 Appendix. 



BOOKS, ETC., ON " THEORETICAL NAVAL 
ARCHITECTURE." 

"Transactions of the Institution of Naval Architects." 

" Transactions of the North-East Coast Institution of Engineers 

and Shipbuilders." 

" Transactions of the Institution of Engineers and Shipbuilders 
in Scotland." 

The papers of these Institutions are usually reproduced in the Engineer- 
ing Journals shortly after the time of the meetings. 

" Transactions of the American Society of Naval Architects." 

The papers of this Society are usually reproduced in the New York 
Journal Marine Engineering, which can now be obtained in this country, 
price 6d. 

" Shipbuilding, Theoretical and Practical." By Prof. Rankine and 
Mr. F. K. Barnes, M.I.N.A. 
A book of great historical interest. 

" Naval Science." Edited by Sir E. J. Reed, K.C.B., F.R.S. 

This was issued for four years, and then discontinued. 
" Theoretical Naval Architecture." By Mr. Samuel J. P. Thearle, 
M.I.N.A. 

This book does not appear to have undergone any revision since its first 
publication. 

" Yacht Architecture." By Mr. Dixon Kemp, Assoc. I.N.A. 

An indispensable volume to those engaged in the design and con- 
struction of yachts. 

" Manual of Naval Architecture." By Sir W. H. White, K.C.B., 
F.R.S. 

The standard treatise on the subject. 
" Stability of Ships." By Sir E. J. Reed, K.C.B., F.R.S. 

Contains much information as to French methods of dealing with 
stability. 

" Text Book of Naval Architecture," for the use of Officers of the 
Royal Navy. By Prof. J. J. Welch, M.I.N.A. 

A very useful text-book on construction, etc., of war vessels. 



Appendix. 489 

" Know your own Ship," for the use of Ships' Officers, etc. By 
Mr. Thomas Walton. 

A most valuable book ; although written for ships' officers, the student 
of Naval Architecture will find much useful information. 

" Naval Architects', Shipbuilders', and Marine Engineers' Pocket 
Book." By Mackrow and Woollard. 
The latest edition has been completely remodelled. 

" The Marine Engineer's Pocket Book." By Messrs. Seaton and 
Rounthwaite. 

Contains much useful information in Naval Architecture. 

"The Speed and Power of Ships," a Manual of Marine Propulsion 
(2 volumes, plates and text). By Mr. D. W. Taylor, M.I.N.A. 
A standard book. 

" Resistance and Propulsion of Ships." By Professor Durand. 
A systematic treatise by an American professor. 

" Applied Mechanics " (Appendix on " Resistance and Propulsion 
of Ships "). By Professor Cotterill, F.R.S. 
This appendix is worth consulting. 

"Encyclopaedia Britannica" (nth edition), article on "Ship- 
building." By Sir Philip Watts, K.C.B., F.R.S. 

This article is of great value. It is specially rich in diagrams and 
illustrations and up-to-date information about many types of ships, also 
some hitherto unpublished matter on resistance. 

Marine numbers of " Cassier's Magazine," August, 1897 ; November, 
1908. 
These numbers form books of permanent value. 

" The Design and Construction of Steam Ships." By Professor 
Biles, LL.D. 

The first volume deals with ship calculations and strength. The 
second volume deals with stability, waves, resistance and propulsion. The 
third volume, to be issued, will deal with design. 

"The Marine Steam Engine." By the late R. Sennett, R.N., 
and Eng. Vice-Adm. Sir H. J. Oram. 

A standard treatise. Contains much valuable information for Naval 
Architects. 

"Address to Mechanical Science Section, British Association, 
1899." By Sir William White, K.C.B., F.R.S. 
A valuable address which is worth reading. 

" Screw Propeller Computer." By Professor McDermott. 

The little book on the Screw Propeller which accompanies this Com- 
puter contains a most succinct account of the principles of the screw 
propeller. 

" Naval Architecture." By Professor Peabody. 

This book is the substance of the author's lectures at the Massachusetts 
Institute of Technology. It is largely based on the French methods. 



49O Appendix. 

" War Ships." By E. L. Attwood, R.C.N.C. 

Specially prepared for officers of the Royal Navy. It, however, forms 
an introduction to Naval Architecture, so far as regards war vessels, 
which may prove useful to students of the subject. 

" Resistance and Power of Steamships." By W. H. Atherton, 
M.Sc., and A. L. Mellanby, M.Sc. 

An excellent little book. It, however, does not deal with propulsion. 
" Presidential Address to Institution of Civil Engineers, 1904." Sir 
W. H. White, K.C.B., F.R.S. 

A most admirable survey of the position of Naval Architecture in the 
year 1904. 

"Aids to Stability." Captain Owen, A.I.N.A. 

Written for officers of the Mercantile Marine. 
" A Complete Class Book of Naval Architecture." By W. J. Lovett. 

Contains many worked-out examples. 

" Marine Propellers." By S. W. Barnaby, M.Inst.CE. 

This is the standard English work on the subject. 
"The Naval Constructor." By G. Simpson, M.I.N.A. 

A pocket-book issued by an American naval architect. Contains a 
mass of useful information. 

" Hydrostatics." By Prof. Greenhill, F.R.S. 

This book treats the subject in a practical manner likely to be of great 
use to students in Naval Architecture. 

" Lloyd's Report on Masting." A masterly survey of problems con- 
nected with sailing vessels. By the late Mr. William John. 
" Shipyard Practice as Applied to Warship Construction." By 
Neil J. McDermaid, R.C.N.C. 

The author of this work was the Instructor on practical ship building 
to naval construction students at Devonport Dockyard. 
" Turbines." By Prof. Biles. 

Gives much information on propellers. 

" The Carriage of Liquid Cargoes." By Captain Little. 
" Unsolved Problems of Shipbuilding." By Dr. Elgar. 

Being the " Forrest " Lecture before the Institution of Civil Engineers 
for 1907. 
" Steamship Coefficients, Speeds and Powers." By Mr. Fyfe. 

An exhaustive collection of data. 
" Resistance and Propulsion." By Prof. Dunkerley. 

The author of this work was formerly Professor at the Royal Naval 
College, where he lectured on this subject. 

"Ship Construction and Calculations." By G. Nichol, Lloyd's 
Surveyor. 

Includes matter relating to the revised Lloyd's Rules. 
" Structural Design of Warships." By Professor Hovgaard. 

Based on lectures delivered at the Massachusetts Institute of Technology. 
" Ship Form Resistance and Screw Propulsion." By G. S. Baker, 

Superintendent of the William Froude Experimental Tank. 
" The .Strength of Ships." By A. J. Murray. 



INDEX 



ALGEBRAIC expression for area of a 
curvilinear figure, 14 

Amsler's integrator, 197, 419 

Angles, measurement of, 90 

Area of circle, 4 

figure bounded by a plain curve 

and two radii, 15 

portion of a figure between two 

consecutive ordinates, 12 

rectangle, I 

square, I 

triangle, 2 

trapezium, 3 

trapezoid, 2 

wetted surface, 85, 86 

Atwood's formula for statical sta- 
bility, 175 

Augment of resistance, 339 

BARNES' method of calculating sta- 
bility, 1 88 

Beams, 262, 265, 274 
Bilge keels, effect on rolling, 359 
Bilging a central compartment, 34 

an end compartment, 165 

Blom's mechanical method of cal- 
culating stability, 187 
BM, longitudinal, 145 

, , approximations, 150 

, transverse, 107 

, , approximations, ill 

Books on theoretical naval archi- 
tecture, 488 

Brown's Displacement Sheet, 414 
Buoyancy, centre of, 63, 64 
, strains due to unequal distri- 
bution of weight and, 268 
Butt fastenings, strength of, 235, 
239 



Butt straps, treatment of, Admiralty 
and Lloyd's, 239 

CALCULATION of weights, 224 

Captain, stability of, 176 

Cavitation, 341 

Centre of buoyancy, 63, 64 

, approximate position, 

65 

of flotation, 98 

of gravity, 47 

of an area bounded by a 

curve and two radii, 60 

of an area with respect to 

an ordinate, 53, 57 

of an area with respect to 

the base, 58 

of a plane area by experi- 
ment, 51 

of a ship, calculation of, 

231 

of outer bottom plating, 

232 

of solid bounded by a 

curved surface and a plane, 62 

of solids, 52 

Change of trim, salt to river, 169 

Circle, area of, 4 

Circular measure of angles, 90 

Coefficient of fineness, displace- 
ment, 31, 32 

midship section, 29 

water-plane, 31 

speed, 315 

Combination table for stability, 193 

Comparison, law of, 321 

Conditions of equilibrium, 92 

stable equilibrium, 96 

Corresponding speeds, 319 



4Q2 



Index. 



Crank ship, 127 

Cross-curves of stability, 197,413 

Cubes and squares, 438 

Curve of areas of midship section, 29 

of bending moment, 271-273 

of buoyancy, 133 

of displacement, 24 

of flotation, 133 

of loads, 271-273 

of sectional areas, 21 

of shearing force, 271-273 

of stability, 176, 182, 183 

, calculation of, 186 

tons per inch immersion, 29 

Curves, use of, in ' calculating 
weights, 228 

DAVITS, strength of, 241 

Derricks, strength of, 245 

Difference in draught, salt and 
river water, 32 

Direct method of calculating sta- 
bility, 196 

Displacement, 23 

, curve of, 24 

of vessel out of the designed 

trim, 152 

sheet, 69, 410, 414 

Draft aft remaining constant, 163 

, change of, due to different 

density of water, 32 

Draught when launched, 170 

Dynamical stability, 204 

EDDY-MAKING resistance, 302, 306 

Edgar> trial in shallow water, 328 

Effective horse-power, 297 

calculation of, 331 

English's, Col., model experiments, 
328 

Equilibrium, conditions of, 92 

, stable, conditions of, 96 

Equivalent girder, 281 

Examination of the Board of Edu- 
cation, questions, 450 

, syllabus, 446 

Experimental data as to strength of 
plates and rivets, 237 

Experiments on Greyhound, 298 

to determine frictional resist- 
ance, 303 

FIVE-EIGHT rule, 12 

Floating dock, information for, 171 



Framing, weight of, 228 
Free water in a ship, 128 
Frictional resistance, 302, 303 
Froude, Mr., experiments of, 298, 
303 

GM by experiment, 119 

GM, values of, 125 

Graphic integration of rolling equa- 
tion, 367 

method of calculating dis- 
placement and position of C.B., 
76 

Greyhound, H.M.S., experiments 
on, 298 

Grounding, loss of stability, 416 

Gun fire, heel produced by, 215 

HOGGING strains, 261 

Hok's method of calculating sta- 
bility, 208 

Horse-power, 296 

, effective, 297 

, indicated, 300 

Hull efficiency, 339 

, weight of, 229 

INCLINING experiment, 119 
Indicated horse-power, 300 
Inertia, moment of, 101 
Integraph, 272 

Integrator, Amsler's, 197, 419 
Interference between bow and stern 

series of transverse waves, 312 
Iron, weight of, 37, 38 

LAUNCHING, calculations for, 400 

draught, 1 70 

Leclert's theorem, 137 
Lifeboats, stability of, 210 
Lloyd's numbers for regulating 

scantlings, 230 
Logarithms, table of, 434 
Longitudinal bending strains, 258, 

265 

BM, 145 

C.G. of a ship, 234 

metacentre, 144 

metacentric height, 1 54 

MATERIALS for shipbuilding, weight 

of, 37 
Measured-mile trials, 326 



Index. 



493 



Mechanical method of calculating 

stability, 187 
Metacentre, longitudinal, 144 

, transverse, 94 

Metacentric diagram, 113 

for simple figures, 132 

height by experiment, 1 19 

when inclined about another 

axis, 217 

, values of, 125 

Miscellaneous examples. 421 
Moment of an area about a line, 

52 
Moment of inertia, 101 

of curvilinear figure, 105 

, approximation to, 

1 06 
Moment to change trim one inch, 

155 
, approximate, 156, 

173 (Ex. 18) 

Monarch, stability of, 177 
Morrish's formula for position of 

C.B.,65 
Moseley's formula for dynamical 

stability, 205 

NORMAND'S approximate formula 
for longitudinal BM, 150 

OUTER bottom plating, weight of, 
228 

PANTING, 258, 264 

Pillars, 244 

Pitch, 343 

Planimeter, 81 

Preliminary table for stability, 192 

Principal stress, 284 

Prismatic coefficient of fineness, 32 

Progressive speed trials, 309 

Propulsion, 337 

Propulsive coefficient, 301 

QUESTIONS set in examination of 
the Board of Education (late the 
Science and Art Department), 
45 

RACKING strains, 263 

Rectangle, area of, I 

Reserve of dynamical stability, 214 

Resistance, 302 

Rolling of ships, 348 



Rolling, strains due to, 263, 353 
Rudder-head, strength of, 395 

SAGGING strains, 261 
Sailing ships, stability of, 213 
Shaft brackets, form of, 306 

, strength of, 250 

Shearing stresses, 282 

Sheer drawing, 69 

Shift of G.C. of a figure due to shift 

of a portion, 100 
Simpson's first rule, 6 
, approximate proof, 

8 

, proof, 407 

second rule, 10 

proof, 408 

Sinkage due to bilging a central 

compartment, 34 

Slip, 343 

"Smith" correction, 285 
Space passed over by ship, 334 
Speed, coefficients of, 315 
Squares and cubes, 438 
Stability, curves of, specimen, 182 

dynamical, 204 

, Moseley's formula, 205 

statical, 93 

, at large angles, 174 

, cross-curves of, 197, 413 

, curve of, 176 

, calculations for, 186 

, definition, 93 

Steadiness, 127 

Steel, weight of, 37 

Stiffness, 126 

Strains experienced by ships, 258 

Strength of butt fastenings, 235 

Stress on material composing the 

section, 273 

Subdivided intervals, 13 
Submerged body, resistance of, 314 
Syllabus of examinations of the 
Board of Education (late the 
Science and Art Department), 
446 

TANGENT to cross curve, 203 
to curve of centres of buoy- 
ancy, 1 1 8, 136 

of metacentres, 137 

curve of stability at the origin, 

182 
Tchebycheffs rules, 17, 409 



494 



Index. 



Tensile tests for steel plates, 

Admiralty, 239 

, Lloyd's, 239 

Thrust deduction, 339 
Timber, weight of, 37 
Tons per inch immersion, 28 
Transverse BM, 107 

metacentre, 94 

strains on ships, 263 

Trapezium, area of, 3 

, C.G. of, 50 

Trapezoid, area of, 2 

Trapezoidal rule, 5 

Trochoidal wave, construction of, 

285 

theory, 285, 289 

Triangle, area of, 2 

, C.G. of, 50 

Trigonometry, 90 
Trim, change of, 153 

moment to change, 155 

Turning, heel due to, 216 



Turning of ships, 381 
UNSYMMETRICAL bending, 284 

VELOCITY of inflow of water, 37 
Volume of pyramid, 19 

of rectangular block, 19 

of solid bounded by a curved 

surface, 20 
Volume of sphere, 19 

WAKE, 337 

Water, free, effect on stability, 128 
Wave, internal structure, 285 
Wave-making resistance, 308, 313 
Weight, effect on trim due to adding, 
159, 161 

of hull, 229 

of materials, 38 

of outer bottom plating, 228 

of steel angles, 225 

Wetted surface, area of, 85, 86 
Wood, weight of, 38 



THE END. 



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n 194 

rngi 







10m-7,'44(1064s) 



840838 






Bering 
( library 



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